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A same length is measured 1.00 m, 3.28 ft, e.t.c., so what?
Topics
About:
elementary school mathematics
The table of contents of this article
Starting Context
Target Context
-
The reader will know the relation between the reality and measurements.
Orientation
There is an article on becoming a benefactor of humanity by being a conduit of truths
Main Body
1: 'Measurement' Is a Basis for Applying Mathematics to the Reality
Special-Student-7-Hypothesizer
As I look at a Japanese elementary school mathematics curriculum, children 1st learn natural numbers with their additions, subtractions, and multiplications, and then, their divisions together with fractions and decimal fractions, but intermingled with the process, children learn measurements.
Special-Student-7-Rebutter
Theoretically speaking, learning 1st collections (and maybe sets), the natural numbers set, the integral numbers set, the rational numbers set, and the real numbers set seems logical, but the curriculum does not go that way.
Special-Student-7-Hypothesizer
In fact, children do not learn negative numbers until junior high school, and as far as I judge from the curriculums for elementary school, junior high school, and high school that I am looking at right now, children do not seem to learn collections until high school.
Special-Student-7-Rebutter
Really? How could children learn functions or probabilities without the concept of collection?
Special-Student-7-Hypothesizer
I do not know; I am looking only at the curriculums not at any textbook, so, the concept of collection may be being cursorily introduced somewhere mingled in a topic.
Anyway, I wanted to say that 'measurement' is being introduced at the 1st stage of mathematics educations.
Special-Student-7-Rebutter
Well, 'measurement' does not seem any indispensable part of mathematics, or is 'measurement' really a part of mathematics?
Special-Student-7-Hypothesizer
We are not talking about the measure theory in mathematics, which is about measuring some subsets of a set, while 'measurement' here is about measuring the reality, which means appointing a number to the reality.
Special-Student-7-Rebutter
'measurement' here does not seem any part of pure mathematics, but is a basis of applying mathematics to the reality.
Special-Student-7-Hypothesizer
The reason why 'measurement' is learned at the 1st stage is that the school mathematics is not meant to teach pure mathematics but to teach applied mathematics.
Children learn natural numbers in order to count some apples (counting is a kind of measurement) and learn decimal fractions in order to measure the length of a side of a desk.
So, school mathematics does not follow the logical order of the construction of pure mathematics but jump to some urgent applications and conveniently pick up some mathematical entities necessary for the applications.
2: Measurement Is to Represent an Aspect of the Reality by a Mathematical Entity
Special-Student-7-Hypothesizer
In general, 'measurement' is to represent an aspect of the reality by a mathematical entity (typically, a number).
For example, for the reality in which a box contains some apples, a measurement gives 5.
I said "an aspect", because there are many boxes and many non-boxes in the reality and the measurement looked at only the specific box and cared only about the quantity of what are inside the box: we cannot exactly represent the whole reality by anything.
So, not only measuring some lengths by some rulers and measuring some weights by some scales but also countings are measurements.
3: Let Us Measure a Length
Special-Student-7-Hypothesizer
Let us measure the length of a side of a desk.
Special-Student-7-Rebutter
Do we need to?
Special-Student-7-Hypothesizer
Ah, that may sound like just a guff, but is really an important question; certainly, we do not need to measure the length, at least, as far as the desk is concerned: the desk exists without being measured, or more generally, the reality exists without being measured. We will measure the length just because that suits our convenience.
Special-Student-7-Rebutter
I am making sure of that point, because there are some people who are claiming "There is no reality except measurements.".
Special-Student-7-Hypothesizer
According to that claim, the desk did not exist before the human measured the desk, and at the instant the human measured the desk, the desk popped up in the reality, and furthermore, at the instant the human abandoned measuring the desk, the desk puffed out from the reality, ..., is that any serious talk?
Special-Student-7-Rebutter
Those people are adamant.
Special-Student-7-Hypothesizer
... Let us talk about it later.
Anyway, we use a ruler to measure the length.
Special-Student-7-Rebutter
Or a tape measure.
Special-Student-7-Hypothesizer
Yes, but that is not any issue here, so, let us call it a ruler.
An issue is that the ruler may be a meter-ruler, an yard-ruler, or any some-other-unit-ruler.
Accordingly, the measurement may be 0.91 (m), 1.00 (yard), or something.
Special-Student-7-Rebutter
With an error.
Special-Student-7-Hypothesizer
Yes, with an error, because the ruler is not so accurate (the production is not so perfect) and is deformed somewhat by temperature, moisture, gravity, e.t.c., your eyesight is not so accurate, e.t.c..
Students learn that it is crucial to specify the possible error in any measurement.
Anyway, the measurement depends on the unit you choose.
So, measurements are relative with respect to units.
But that does not mean that the reality is relative.
In fact, if you have changed from meter to yard, the measurement has changed, but the length is not changed just because you have chosen a fancy unit.
Special-Student-7-Rebutter
The seed of misleading is the definition of 'length'.
Special-Student-7-Hypothesizer
To simplify the argument, let us forget Relativity throughout this section.
Then, I mean by 'length' how much the 2 points are separated in the space in the reality.
Just because you have used a yard-ruler, the 2 points have not become less separated.
In fact, if you thought that the desk had shrunk just because you used a yard-ruler, you would be crazy; or when a man lives 1,000.000 km separated from his love, if he thought that adopting a unit, perhaps called "yearn", by which he was 0.000,001 yearn separated from his love, would shrink the distance to his love from him, he would be a romanticist.
Likewise, measurements' having some errors does not mean that the reality has some errors.
If you thought that the reality had an error just because the production of the ruler was not perfect and your eyesight was not accurate, you would be crazy, and egocentric, in addition.
Special-Student-7-Rebutter
I talked about "The seed of misleading", because if someone meant by "length" 'measurement of the length', "lengths" would be relative with respect to units.
Special-Student-7-Hypothesizer
We mean by "length" 'how much the 2 points are separated in the space in the reality', and when we mean 'measurement of the length', we use "measurement of the length".
In fact, calling 'measurement of the length' "length" is a sloppy abbreviation, and some people refuse to refrain from using such sloppy abbreviations, which causes confusions.
When a measurement of the length of a side of a desk is 1.00 m and a measurement of the length of a side of another desk is 1.00 yard, we do not say "The 2 desks have the same length.".
4: But How Does Relativity Change the Situation?
Special-Student-7-Rebutter
It is good so far for non-Relativistic cases, but how does Relativity change the situation?
Special-Student-7-Hypothesizer
Relativity does not say "the reality is relative" at all, although it is rather prevalently misunderstood to be saying so.
Special-Student-7-Rebutter
What does Relativity say about "the length of the side of the desk"?
Special-Student-7-Hypothesizer
Relativity says that the universe is a 4-dimensional spacetime.
Note that the spacetime itself is not relative in any way.
But when you begin to talk about a "space", what is the "space"?
Special-Student-7-Rebutter
The "space" seems to mean a 3-dimensional space.
Special-Student-7-Hypothesizer
When you talk about a 3-dimensional "space", you mean a cross section of the 4-dimensional spacetime.
But you can choose infinitely many cross sections of the 4-dimensional spacetime.
So, the "space" depends on your choice, which means that "space"s are relative with respect to your choices.
If someone finds difficult to imagine the 4-dimensional spacetime, he or she could imagine the spacetime as a 2-dimensional plane, as a metaphor.
Then, a "space" would be a line of your choice on the plane.
The 2 ends of the side of the desk draw 2 so-called "world line"s in the spacetime, and when you choose a "space", each of the world lines intersects the "space" at a point, and when you say "how much the 2 points are separated in the space in the reality", "the 2 points" are the 2 intersections.
As the 2 intersections depend on your choice of the "space", the length depends on your choice of the "space", which is what Relativity says.
Special-Student-7-Rebutter
Whatever "space" you choose, that does not change the reality at all, which is parallel to that choosing a unit of length does not change the reality.
Special-Student-7-Hypothesizer
'choosing a "space"' is typically related with choosing a 3-dimensional coordinates system: the 3-dimensional coordinates system at a "time" is really a cross-section of the 4-dimensional spacetime, which is called "space", and the 4-dimensional spacetime is sliced into such parallel "space"s with different "time"s.
And some 2 3-dimensional coordinates systems moving against each other have sliced the 4-dimensional spacetime in some different directions.
Special-Student-7-Rebutter
Is it not bad to adopt a 3-dimensional coordinate system for the 4-dimensional reality?
Special-Student-7-Hypothesizer
In fact, it is misleading if not absolutely bad: people tend to imagine as though "the space" is developing through "time", but your "space" is just your choice, not any objective existence.
Now, you should understand why the 2 coordinates systems that are moving against each other measure the side of the desk differently.
In the 2-dimensional spacetime metaphor, the 2 ends of the side of the desk draw some parallel world lines on the 2-dimensional plane (let the plane be the x-y plane and let the 2 world lines be x = 1 and x = 2), and the 1st coordinates system is the bunch of y = t lines and the 2nd coordinates system is the bunch of y = x + t' lines.
Then, the 1st coordinates system sees at t = 0, the 2 ends of the side of the desk at (1, 0) and (2, 0), while the 2nd coordinates system sees at t' = 0, the 2 ends of the side of the desk at (1, 1) and (2, 2).
So, the 1st coordinates system talks about the length between (1, 0) and (2, 0) and the 2nd coordinates system talks about the length between (1, 1) and (2, 2), and it is not mysterious that they report some different measurements, because the 2 coordinates systems are talking about the different pairs of points.
Special-Student-7-Rebutter
As a caveat, the 4-dimensional spacetime has the Lorentzian metric, not the Euclidean metric.
Special-Student-7-Hypothesizer
Yes, "the Euclidean metric" means that the length between \((x_1, x_2, x_3, x_4)\) and \((x'_1, x'_2, x'_3, x'_4)\) is \(\sqrt{(x'_1 - x_1)^2 + ... + (x'_4 - x_4)^2}\), but "the Lorentzian metric" does not say so (we will not discuss here what exactly it says).
So, in the 2-dimensional metaphor, the length between (1, 0) and (2, 0) is not necessarily 1 and the length between (1, 1) and (2, 2) is not necessarily \(\sqrt{2}\).
So, we are not saying "the length between (1, 1) and (2, 2) is larger than the length between (1, 0) and (2, 0)" but are saying just that it is natural that the 2 coordinates systems report different measures.
Special-Student-7-Rebutter
But the reality as the 2-dimensional plane is not concerned with what coordinates systems a crazy human has chosen or what measurements are with respect to such coordinates systems.
Special-Student-7-Hypothesizer
As some 2 rulers with different units measure the desk differently, some 2 rulers moving against each other measure the desk differently, which is no mystery: it is about that just like the 2 rulers with different units are different rulers, the 2 rulers moving against each other are different rulers, and it is natural that some different rulers measure differently.
In fact, an observer (a coordinates system) flies by the earth with a high speed relative to the earth and measures the length between the 2 lovers as 1 cm, but how is that a consolation for the poor romanticist?
5: The Reality Is No Uncertain
Special-Student-7-Hypothesizer
Another prevalent misunderstanding is "the reality is uncertain, by the Heisenberg uncertainty principle".
No.
The Heisenberg uncertainty principle is claiming that any simultaneous measurement of each of some pairs of observables is uncertain.
Special-Student-7-Rebutter
What are uncertain are some measurements not the reality.
Special-Student-7-Hypothesizer
And there is also a prevalent misunderstanding that the break of Bell's inequality has proved "the uncertainty of the reality".
No.
The break of Bell's inequality is also suggesting the uncertainty of some measurements.
Special-Student-7-Rebutter
We need to distinguish the reality and its measurements.
Special-Student-7-Hypothesizer
The reality exists without being measured and does not need to be measured.
For example, did the Big Bang need to be measured for it to exist?
Was the Big Bang uncertain until a human measured it?
Special-Student-7-Rebutter
Has the Big Bang been measured even now?
Special-Student-7-Hypothesizer
It seems impossible to fully measure the Big Bang, while some humans have measured a tiny part of the aftermath of the Big Bang.
So, does the Big Bang become somewhat a little more certain at the instant a human measures a tiny part of the aftermath, retroactively?
Special-Student-7-Rebutter
In fact, any measurement is really an aftermath of what was intended to be measured in the reality.
Special-Student-7-Hypothesizer
When a human tries to measure an aspect of the reality, the human puts an apparatus near the aspect, and the apparatus and the aspect interact somehow, and the result of the interaction is the measurement.
If you claim that the aspect is determined retroactively because of the interaction, that is a violation of causality.
Special-Student-7-Rebutter
I thought that even quantum mechanics did not refute causality.
Special-Student-7-Hypothesizer
If the laws that govern the reality are probabilistic, that interaction is probabilistic, and the measurement is not the exact representation of the aspect.
Bell's inequality is based on the assumption that measurements are the exact representations of the aspects of the reality, and its break is suggesting that the assumption is wrong.
Note that Bell's inequality was introduced for the purpose of supporting Mr. Einstein's claim, "God does not play at dice", and naturally supposed that the laws of the reality were deterministic, which would imply that measurements as interactions were deterministic, which the results seem to have refuted.
Although there are some prevalent misstatements like "The reality is not determined until it is measured.", absolutely no, 'measurements are not determined until the measurements are performed' seems a more correct statement, which is natural because any measurement is a physical phenomenon and as far as the laws of the reality are probabilistic, the outcome of the measurement is not determined until the outcome happens.
6: Measurements Are Harassments
Special-Student-7-Hypothesizer
Let us think of an allegory.
You, an Earthian, are captured by an alien species.
The alien species feel only 'fear' and 'anger', and so, they understand only 'fear' and 'anger', in fact, they cannot ever imagine whatever else except 'fear' and 'anger' exist in the reality.
You feel 'love', 'pity', 'joy', 'sadness', e.t.c., I hope.
The alien species measure you, which is to determine whether you are in 'fear' or in 'anger', for the alien species, because that is all they can ever imagine to exist.
So, the alien species harass you to force you to show 'fear' or 'anger', and they are satisfied when you show 'fear' or 'anger' saying "Oh, this maggot was in fear!".
They invent Bell's inequality and conclude that your state was not determined until they measured you, and they even begin to claim "We determine the reality!".
Special-Student-7-Rebutter
Of course, you were not in fear until they measured (harassed) you.
Special-Student-7-Hypothesizer
The alien species's conceptions are only "in fear or in anger", and when you are in pity, they declare "This maggot's state is uncertain." and claim "The reality is not determined until we measure it.".
Special-Student-7-Rebutter
"state" by them is the state brought about by their harassment, and certainly, the "state" does not appear in the reality until they perform the harassment.
Special-Student-7-Hypothesizer
And if your reaction to the harassment is probabilistic, the "state" was certainly not determined until they performed the measurement.
But that does not mean you had not been in an absolutely definite state of pity before the harassment was performed.
Special-Student-7-Rebutter
We should be aware that human concepts like "position", "spin", e.t.c. are derived from pathetically limited human daily experiences, and Earthians who are trying to measure the reality in such concepts are like the alien species who are trying to measure you in 'fear' or 'anger'.
7: Isn't There a Misunderstanding About 'Scientific Method'?
Special-Student-7-Hypothesizer
Isn't there a misunderstanding about 'scientific method'?
Certainly, science values and requires measurements, because just willfully claiming something without any measurement does not give any credibility.
But science is also based on the assumption of the existence of the objective reality.
In fact, if there was no objective reality, what would we measure?
I thought that as we supposed the existence of the objective reality, we were trying to measure the reality. If measurements are relative and uncertain, that is not convenient for us, but we must accept it, and we try to approach the objective reality as near as possible, if not we reach the objective reality, through trial-and-errors of making hypotheses and checking them with measurements, which is the scientific method, I thought.
The scientific method is never "There is no reality except measurements.", I thought.
Special-Student-7-Rebutter
If "There is no reality except measurements", why don't we just do willful measurements to get desired results? We would not need to pay any respect to non-existent reality, so, any willful measurement would be as good as any measurement.
Special-Student-7-Hypothesizer
'accuracy' is the difference between the objective reality and the measurement, and if there was no objective reality, we would not need to care about any accuracy, because there would not be such thing as 'accuracy', because there would be not anything with which the measurement is compared for accuracy.
So, we would concoct an apparatus that would give '1' anyway, and we would find a golden theory, "Measurements are always '1'.". ... Why not? As only we were creating the reality, why wouldn't we create the willful reality?
Or you should just stop any measurement, and there would be no reality, and "There is no reality." would be the absolutely complete theory.
Special-Student-7-Rebutter
Sun will burn out in short time, Earthians will perish with it, and just because pathetic Earthians will perish on a pathetic remote tiny planet, will the reality disappear because measurements will not be performed anymore?
8: A Bad Moral Implication
Special-Student-7-Hypothesizer
In fact, "There is no reality except measurements." is the root of every evil.
"More people came to my inauguration than to the predecessor's!" would be an "alternative truth", because that perception was a measurement and there would be no objective reality to refute the measurement.
There would be no lie, because there would be no objective reality to compare with for the accuracy of the statement.
So, a villain would claim anything willfully with indemnity.
Sounds familiar?
Blackguards cite some misleading YouTube videos or TV programs that claim that Relativity and Bell's inequality have proved "the relativity and uncertainty of the reality" to justify themselves, saying "So, there is no objective truth, so, whatever I say are as truthful as whatever anyone says.".
...
Instead, we are supposed to accept the existence of the objective reality and be humble that our present claim is somehow relative and uncertain, so, we need to be ready to listen to claims by others to remedy our claim to make it more impartial.
Special-Student-7-Rebutter
Relativity and Bell's inequality are in fact some opportunities for us to be humble if they are understood properly, but with their understood improperly (via improper promotions), they are enticing absolute arrogance.
9: Conclusion
Special-Student-7-Hypothesizer
After all, what good children need to engrave in their minds at this point is the relation between the reality and measurements.
The reality exists independent of any measurement and measurements are relative and uncertain, without that presumption, any science, including not only natural sciences but also social sciences, would become non-science.
Special-Student-7-Rebutter
While there are many people who claim that they do not need to study mathematics ("because mathematics is not any useful for their daily lives", they say), there are some things understood best (or probably only) via mathematics.
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series |
definition of left-invariant vectors field over Lie group
Topics
About:
group
About:
\(C^\infty\) manifold
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of left-invariant vectors field over Lie group.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( G\): \(\in \{\text{ the Lie groups }\}\)
\(*V\): \(\in \{\text{ the vectors fields over } G\}\)
//
Conditions:
\(\forall g \in G ((V, V) \in \{\text{ the } l_g \text{ -related vectors fields pairs }\})\), where \(l_g: G \to G, g' \mapsto g g'\) is the left-translation by \(g\)
//
2: Note
\(V\) is inevitably \(C^\infty\), by the proposition that any left-invariant vectors field over any Lie group is \(C^\infty\).
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for topological space and subspace with some connected-components removed, remained connected-components are connected-components of subspace
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any topological space and the subspace with any some connected-components removed, the remained connected-components are the connected-components of the subspace.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(S'\): \(= \{C_j \in \{\text{ the connected components of } T'\} \vert j \in J\}\)
\(S\): \(\subset S'\)
\(T\): \(= \cup S \subset T'\), with the subspace topology
//
Statements:
\(S = \{\text{ the connected components of } T\}\)
//
2: Proof
Whole Strategy: Step 1: see that each \(C_j \in S\) is connected on \(T\); Step 2: see that for each \(j \neq l\), each point in \(C_j\) is not connected to any point in \(C_l\) on \(T\); Step 3: conclude the proposition.
Step 1:
Each \(C_j \in S\) is connected on \(T\), by the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
Step 2:
Let us see that for each \(j, l \in J\) such that \(j \neq l\), each point in \(C_j\) is not connected to any point in \(C_l\) on \(T\).
Let us suppose that a \(c_j \in C_j\) was connected to a \(c_l \in C_l\) on \(T\).
There would be a connected subspace of \(T\), \(T^` \subseteq T\), such that \(c_j, c_l \in T^`\).
\(T^`\) would be a connected subspace of \(T'\), by the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace, which would mean that \(c_j\) and \(c_l\) were connected on \(T'\), a contradiction against that \(c_j\) and \(c_l\) were not in the same connected-component of \(T'\).
So, for each \(j, l \in J\) such that \(j \neq l\), each point in \(C_j\) is not connected to any point in \(C_l\) on \(T\).
Step 3:
So, each \(C_j\) is a connected subspace of \(T\) that cannot be made larger, so, \(C_j\) is a connected component of \(T\), by the proposition that any connected topological component is exactly any connected topological subspace that cannot be made larger.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for topological space and set of connected subspaces, subspace is contained in connected component of union of subspaces
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any topological space and any set of connected subspaces, each subspace is contained in a connected component of the union of the subspaces.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T_j \vert j \in J\}\): \(T_j \in \{\text{ the connected topological subspaces of } T'\}\)
\(T\): \(= \cup_{j \in J} T_j \subseteq T'\) with the subspace topology
//
Statements:
\(\forall j \in J (\exists C \in \{\text{ the connected components of } T\} (T_j \subseteq C))\)
//
2: Proof
Whole Strategy: Step 1: see that \(T_j\) is a connected subspace of \(T\); Step 2: conclude the proposition.
Step 1:
\(T_j\) is a connected subspace of \(T\), by the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
Step 2:
Let us take the union of all the connected subspaces of \(T\) that contain \(T_j\), which is a connected subspace of \(T\) that cannot be made larger, by the proposition that for any topological space, the union of any connected subspaces that share any point is connected: that cannot be made larger because if there was a larger one, the larger one would be contained in the union.
So, by the proposition that any connected topological component is exactly any connected topological subspace that cannot be made larger, the union is a connected component of \(T\) in which \(T_j\) is contained.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for locally connected topological space, open subspace is locally connected
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any locally connected topological space, any open subspace is locally connected.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T'\): \(\in \{\text{ the locally connected topological spaces }\}\)
\(T\): \(\in \{\text{ the open subsets of } T'\}\) with the subspace topology
//
Statements:
\(T\): \(\in \{\text{ the locally connected topological spaces }\}\)
//
2: Note
As Proof shows, \(T\) needs to be open in order for this proposition to be applied.
3: Proof
Whole Strategy: Step 1: for each \(t \in T\), take any neighborhood of \(t\) on \(T\), \(N'_t\), and see that \(N'_t\) is a neighborhood of \(t\) on \(T'\); Step 2: take a connected neighborhood of \(t\) on \(T'\) contained in \(N'_t\), \(N_t\); Step 3: see that \(N_t\) is a connected neighborhood of \(t\) on \(T\).
Step 1:
Let \(t \in T\) be any.
Let \(N'_t \subseteq T\) be any neighborhood of \(t\).
\(N'_t\) is a neighborhood of \(t \in T'\) on \(T'\), because there is an open neighborhood of \(t\) on \(T\), \(U'_t \subseteq T\), such that \(U'_t \subseteq N'_t\), which is open also on \(T'\), by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
Step 2:
There is a connected neighborhood of \(t\) on \(T'\), \(N_t \subseteq T'\), such that \(t \in N_t \subseteq N'_t\), by the definition of locally connected topological space.
Step 3:
\(N_t \subseteq N'_t \subseteq T\) is a neighborhood of \(T\), because there is an open neighborhood of \(t\) on \(T'\), \(U_t \subseteq T'\), such that \(U_t \subseteq N_t\), which is an open neighborhood of \(t\) also on \(T\), by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
While \(N_t\) is a connected subspace of \(T'\), \(N_t\) is a connected subspace of \(T\), by the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
So, \(N_t\) is a connected neighborhood of \(t\) on \(T\).
So, \(T\) is locally connected.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for topological space and \(2\) connected subspaces, connected-components of union of subspaces are subspaces or union of subspaces
Topics
About:
topological space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any topological space and any \(2\) connected subspaces, the connected-components of the union of the subspaces are the subspaces or the union of the subspaces.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(T_1\): \(\in \{\text{ the connected subspaces of } T'\}\)
\(T_2\): \(\in \{\text{ the connected subspaces of } T'\}\)
\(T_1 \cup T_2\): \(\subseteq T'\) as the topological subspace
//
Statements:
\(T_1 \cup T_2\) has the connected-components, \(\{T_1, T_2\}\) or \(\{T_1 \cup T_2\}\)
//
2: Note
Refer to the proposition that for any topological space, the union of any 2 connected subspaces is connected if each neighborhood of a point on one of the subspaces contains a point of the other subspace and the proposition that for any topological space, the union of any connected subspaces that share any point is connected.
3: Proof
Whole Strategy: Step 1: see that for each \(t_1, t'_1 \in T_1\), \(t_1\) and \(t'_1\) are connected on \(T_1 \cup T_2\) and for each \(t_2, t'_2 \in T_2\), \(t_2\) and \(t'_2\) are connected on \(T_1 \cup T_2\); Step 2: suppose that each \(t_1\) is not connected to any \(t_2\), and see that \(\{T_1, T_2\}\) are the connected components; Step 3: suppose that a \(t_1\) is connected to a \(t_2\), and see that \(\{T_1 \cup T_2\}\) is the connected component.
Step 1:
Let \(t_1, t'_1 \in T_1\) be any.
\(T_1\) as the subspace of \(T'\) is the subspace of \(T_1 \cup T_2\), by the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
So, \(T_1\) is a connected subspace of \(T_1 \cup T_2\) that contains \(t_1\) and \(t'_1\).
So, \(t_1\) and \(t'_1\) are connected on \(T_1 \cup T_2\).
For each \(t_2, t'_2 \in T_2\), \(t_2\) and \(t'_2\) are connected on \(T_1 \cup T_2\).
Step 2:
There are only the 2 cases: 1) each \(t_1 \in T_1\) is not connected to any \(t_2 \in T_2\); 2) a \(t_1 \in T_1\) is connected to a \(t_2 \in T_2\).
Let us suppose that 1) each \(t_1 \in T_1\) is not connected to any \(t_2 \in T_2\).
\(T_1\) is an equivalence class of connected-ness on \(T_1 \cup T_2\), because each pair of points on \(T_1\) are connected by Step 1 and each point not on \(T_1\) is on \(T_2\) and does not belong to the equivalence class, by the supposition.
\(T_2\) is an equivalence class of connected-ness on \(T_1 \cup T_2\), likewise.
So, \(\{T_1, T_2\}\) are the connected components of \(T_1 \cup T_2\).
Step 3:
Let us suppose that 2) a \(t_1 \in T_1\) is connected to a \(t_2 \in T_2\).
Each pair of points on \(T_1 \cup T_2\) are connected, because when they are \(t'_1, t''_1 \in T_1\), they are connected by Step 1; when they are \(t'_1 \in T_1\) and \(t'_2 \in T_2\), they are connected because \(t'_1\) and \(t_1\) are connected, \(t_1\) and \(t_2\) are connected, and \(t_2\) and \(t'_2\) are connected; when they are \(t'_2, t''_2 \in T_2\), they are connected by Step 1.
So, \(T_1 \cup T_2\) is the equivalence class of connected-ness on \(T_1 \cup T_2\).
So, \(\{T_1 \cup T_2\}\) is the connected component of \(T_1 \cup T_2\).
References
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description/proof of that for set, union of subsets minus subset is union of (each of former subsets minus latter subset) s
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any set, the union of any subsets minus any subset is the union of (each of the former subsets minus the latter subset) s.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(S'\): \(\in \{\text{ the sets }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \subseteq S' \vert j \in J\}\):
\(S\): \(\subseteq S'\)
//
Statements:
\((\cup_{j \in J} S_j) \setminus S = \cup_{j \in J} (S_j \setminus S)\)
//
2: Proof
Whole Strategy: Step 1: see that for each \(s \in (\cup_{j \in J} S_j) \setminus S\), \(s \in \cup_{j \in J} (S_j \setminus S)\); Step 2: see that for each \(s \in \cup_{j \in J} (S_j \setminus S)\), \(s \in (\cup_{j \in J} S_j) \setminus S\).
Step 1:
Let \(s \in (\cup_{j \in J} S_j) \setminus S\) be any.
\(s \in \cup_{j \in J} S_j\), so, \(s \in S_j\) for a \(j \in J\).
\(s \notin S\).
So, \(s \in S_j \setminus S\) for that \(j\).
So, \(s \in \cup_{j \in J} (S_j \setminus S)\).
Step 2:
Let \(s \in \cup_{j \in J} (S_j \setminus S)\) be any.
\(s \in S_j \setminus S\) for a \(j \in J\).
So, \(s \in S_j\) for that \(j\) and \(s \notin S\).
So, \(s \in \cup_{j \in J} S_j\).
So, \(s \in (\cup_{j \in J} S_j) \setminus S\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that antisymmetric real matrix can be block-diagonalized by orthogonal matrix
Topics
About:
matrices space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that any antisymmetric real matrix can be block-diagonalized by an orthogonal matrix.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(M\): \(\in \{\text{ the } n \times n \text{ real antisymmetric matrices } \}\)
//
Statements:
\(\exists O \in \{\text{ the orthogonal matrices }\} (O^t M O = \begin{pmatrix} 0 & \sqrt{\lambda_1} & 0 & ... & & & & & & & 0 \\ - \sqrt{\lambda_1} & 0 & ... & & & & & & & & 0 \\ 0 & ... & 0 & \sqrt{\lambda_2} & 0 & ... & & & & & 0 \\ 0 & 0 & - \sqrt{\lambda_2} & 0 & ... & & & & & & 0 \\ ... \\ 0 & ... & & && & 0 & \sqrt{\lambda_{2 m}} & 0 & ... & 0 \\ 0 & ... & & & & 0 & - \sqrt{\lambda_{2 m}} & 0 & ... & & 0 \\ 0 & ... & & & & & & & & & 0 \\ ... \\ 0 & ... & & & & & & & & & 0 \end{pmatrix})\), where \(\{\lambda_1, ..., \lambda_{2 m}, 0, ..., 0\}\) are the eigenvalues of \(M^t M\) where \(0 \lt \lambda_j\)
//
2: Note
"block-diagonalized" means that the result has some diagonal blocks (which means blocks at diagonal positions, not blocks having diagonal shapes), each of which is \(\begin{pmatrix} 0 & \sqrt{\lambda_j} \\ - \sqrt{\lambda_j} & 0 \end{pmatrix}\) or \(\begin{pmatrix} 0 \end{pmatrix}\), with the other components \(0\).
Obviously, any nonzero antisymmetric matrix cannot be diagonalized, because the diagonal components of the antisymmetric matrix are all \(0\), so, \(O^t M O\) would be the \(0\) matrix (\(O^t M O\) is antisymmetric), and \(M = O O^t M O O^t = 0\).
3: Proof
Whole Strategy: Step 1: see that \(M^t M\) is symmetric and has the decreasing eigenvalues (with any duplications), \((\lambda_1, ..., \lambda_k, 0, ..., 0)\), where \(0 \lt \lambda_j\), with some eigenvectors, \(e_1, ..., e_n\); Step 2: see that \(M e_j\) is an eigenvector for \(\lambda_j\) orthogonal to \(e_j\), so, \((e_j, M e_j)\) is a pair for the same \(\lambda_j\); Step 3: take an orthonormal eigenvectors of \(M^t M\), \((O_1, ..., O_{2 m}, O_{2 m + 1}, ..., O_n)\), with the eigenvalues, \((\lambda_1, ..., \lambda_{2 m}, 0, ..., 0)\); Step 4: take \(O\) as \(\begin{pmatrix} O_1 & ... & O_n \end{pmatrix}\); Step 5: see that \(O^t M O\) is as is demanded.
Step 1:
\(M^t M\) is symmetric, because \((M^t M)^t = M^t {M^t}^t\), by the proposition that for any commutative ring, the transpose of the product of any matrices is the product of the transposes of the constituents in the reverse order, \(= M^t M\).
So, \(M^t M\) has the eigenvalues (ordered decreasingly for our convenience), \((\lambda_1, ..., \lambda_n)\), with any duplications, with some eigenvectors, \((e_1, ..., e_n)\), as is well known.
Let us see that \(0 \le \lambda_j\) for each \(j \in \{1, ..., n\}\).
\(M^t M e_j = \lambda_j e_j\).
\({e_j}^t M^t M e_j = (M e_j)^t M e_j = \Vert M e_j \Vert^2\), which is non-negative.
But \({e_j}^t M^t M e_j = {e_j}^t \lambda_j e_j = \lambda_j {e_j}^t e_j = \lambda_j \Vert e_j \Vert^2\).
So, \(0 \le \lambda_j \Vert e_j \Vert^2\), which implies that \(0 \le \lambda_j\).
So, \((\lambda_1, ..., \lambda_n) = (\lambda_1, ..., \lambda_k, 0, ..., 0)\) where \(0 \lt \lambda_j\), where the "\(0, ..., 0\)" part does not really exist when \(k = n\).
Step 2:
Let \(j \in \{1, ..., k\}\) be any.
Let us see that \(M e_j\) is an eigenvector for \(\lambda_j\) orthogonal to \(e_j\).
\(M^t M (M e_j) = - M^t (- M) (M e_j) = - (- M) (M^t) (M e_j) = M (M^t M e_j) = M (\lambda_j e_j) = \lambda_j (M e_j)\).
On the other hand, \(M (M e_j) = - (- M M e_j) = - (M^t M e_j) = - \lambda_j e_j \neq 0\), which implies that \(M e_j \neq 0\).
\(e_j = - 1 / \lambda_j M (M e_j)\).
\({e_j}^t (M e_j) = (- 1 / \lambda_j M (M e_j))^t M e_j = - 1 / \lambda_j (M (M e_j))^t M e_j = - 1 / \lambda_j (M e_j)^t M^t M e_j = - 1 / \lambda_j (M e_j)^t \lambda_j e_j = - (M e_j)^t e_j = - ((M e_j)^t e_j)^t\), because the transpose of any scalar is the scalar, \(= - {e_j}^t ((M e_j)^t)^t = - {e_j}^t (M e_j)\), which implies that \({e_j}^t (M e_j) = 0\).
So, \(M e_j\) is an eigenvector for \(\lambda_j\) orthogonal to \(e_j\).
So, \(\{e_j, M e_j\}\) is linearly independent, and \((e_j, M e_j)\) forms a pair of eigenvectors for \(\lambda_j\).
Step 3:
\(O_1 := e_1 / \Vert e_1 \Vert\) is a normal eigenvector for \(\lambda_1\).
Let us take \(O_2 := - 1 / \sqrt{\lambda_1} M O_1\), which is an eigenvector for \(\lambda_1\) orthogonal to \(O_1\) by Step 2.
\({O_2}^t O_2 = (- 1 / \sqrt{\lambda_1} M O_1)^t (- 1 / \sqrt{\lambda_1} M O_1) = 1 / \lambda_1 (M O_1)^t M O_1 = 1 / \lambda_1 {O_1}^t M^t M O_1 = 1 / \lambda_1 {O_1}^t \lambda_1 O_1 = {O_1}^t O_1 = 1\), so, \(O_2\) is a normal eigenvector for \(\lambda_1\) orthogonal to \(O_1\).
Note that \(M O_2 = M (- 1 / \sqrt{\lambda_1} M O_1) = - 1 / \sqrt{\lambda_1} M M O_1 = \sqrt{\lambda_1} O_1\), by Step 2.
If there is no more duplication of \(\lambda_1\), \((\lambda_1, \lambda_2 = \lambda_1)\) will be the duplications of \(\lambda_1\).
Let us suppose that there is another duplication of \(\lambda_1\).
A normal eigenvector, \(O_3\), can be taken to be orthogonal to \((O_1, O_2)\), by the definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.
Then, let us take \(O_4 := - 1 / \sqrt{\lambda_1} M O_3\), a normal eigenvector for \(\lambda_1\) orthogonal to \(O_3\), as before.
Let us see that \(O_4\) is orthogonal also to \(O_1\) and \(O_2\).
For \(j \in \{1, 2\}\), \({O_j}^t O_4 = {O_j}^t (- 1 / \sqrt{\lambda_1} M O_3) = - 1 / \sqrt{\lambda_1} {O_j}^t M O_3 = - 1 / \sqrt{\lambda_1} {O_j}^t {M^t}^t O_3 = - 1 / \sqrt{\lambda_1} (M^t O_j)^t O_3 = - 1 / \sqrt{\lambda_1} (- M O_j)^t O_3 = 1 / \sqrt{\lambda_1} (M O_j)^t O_3\), but \(M O_j\) is a scalar multiple of \(O_1\) or \(O_2\), so, \(= 0\).
And so on, after all, \(\lambda_1\) has some even duplications, \((\lambda_1, \lambda_2 = \lambda_1, ..., \lambda_{2 l - 1} = \lambda_1, \lambda_{2 l} = \lambda_1)\), with the orthonormal eigenvectors, \((O_1, O_2, ..., O_{2 l - 1}, O_{2 l})\).
Doing likewise for each eigenvalue-positive-duplications, we have the eigenvalues, \((\lambda_1, ..., \lambda_{2 m})\) with the orthonormal eigenvectors, \((O_1, ..., O_{2 m})\): any 2 eigenvectors with different eigenvalues, \(O_j, O_l\), are inevitably orthogonal to each other, because \((\lambda_l - \lambda_j) {O_j}^t O_l = \lambda_l {O_j}^t O_l - \lambda_j {O_j}^t O_l = {O_j}^t M^t M O_l - (M^t M O_j)^t O_l = ((M^t M)^t O_j)^t O_l - (M^t M O_j)^t O_l = (M^t M O_j)^t O_l - (M^t M O_j)^t O_l = 0\), which implies that \({O_j}^t O_l = 0\).
For the eigenvalue-0-duplications, we take any orthonormal eigenvectors, by the definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.
So, we have the eigenvalues \((\lambda_1, ..., \lambda_{2 m}, 0, ..., 0)\) with the orthonormal eigenvectors, \((O_1, ..., O_{2 m}, O_{2 m + 1}, ..., O_n)\): any 2 eigenvectors with different eigenvalues, \(O_j, O_l\), are inevitably orthogonal to each other, as before.
Step 4:
Let us take \(O := \begin{pmatrix} O_1 & ... & O_n \end{pmatrix}\).
\(O\) is an orthogonal matrix, because \((O_1, ..., O_n)\) is orthonormal: \((O_1, ..., O_n)\)'s being orthonormal is nothing but \(O^t O = I\).
Step 5:
Let us see that \(O^t M O\) is as is demanded.
For each \(2 m \lt j\), \(M O_j = 0\), because \(M^t M O_j = 0\), so, \({O_j}^t M^t M O_j = 0\), but the left hand side is \((M O_j)^t M O_j = \Vert M O_j \Vert^2\), so, \(\Vert M O_j \Vert^2 = 0\), which implies that \(M O_j = 0\).
Let us see that \((O^t M O)^j_l = {O_j}^t M O_l\).
\((O^t M O)^j_l = (O^t)^j (M O)_l\), where \((O^t)^j\) denotes the \(j\)-th row of \(O^t\) and \((M O)_l\) denotes the \(l\)-th column of \(M O\).
\((O^t)^j = {O_j}^t\).
\((M O)_l = M O_l\).
So, \((O^t M O)^j_l = {O_j}^t M O_l\).
For each \(j = 2 r + 1\) for each \(r \in \{0, ..., m - 1\}\), for \(l = j + 1\), \({O_j}^t M O_l = {O_j}^t \sqrt{\lambda_j} O_j = \sqrt{\lambda_j}\), and for any other \(l\), \({O_j}^t M O_l = 0\), because when \(l \le 2 m\), \(M O_l\) is a scalar multiple of the other in the pair to which \(O_l\) belongs, and when \(2 m \lt l\), \(M O_l = 0\).
For each \(j = 2 r + 2\) for each \(r \in \{0, ..., m - 1\}\), for \(l = j - 1\), \({O_j}^t M O_l = {O_j}^t (- \sqrt{\lambda_j} O_j) = - \sqrt{\lambda_j}\), and for any other \(l\), \({O_j}^t M O_l = 0\), because when \(l \le 2 m\), \(M O_l\) is a scalar multiple of the other in the pair to which \(O_l\) belongs, and when \(2 m \lt l\), \(M O_l = 0\).
For each \(j\) such that \(2 m \lt j\), for each \(l \in \{1, .., n\}\), \({O_j}^t M O_l = 0\), because when \(l \le 2 m\), \(M O_l\) is a scalar multiple of the other in the pair to which \(O_l\) belongs, and when \(2 m \lt l\), \(M O_l = 0\).
That meas that \(O^t M O\) is as is demanded.
References
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