<The previous article in this series | The table of contents of this series |
description/proof of that for continuous real map from interval with closed end, if image of interior is bounded, range is correspondingly equal-or-smaller-or-larger bounded
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 continuous real map from any non-1-point interval with any closed end, if the image of the interior is bounded, the range is correspondingly equal-or-smaller-or-larger bounded.
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:
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\)
\(J\): \(\in \{\text{ the non-1-point intervals of } \mathbb{R}\}\), with any closed end
\(f\): \(: J \to \mathbb{R}\), \(\in \{\text{ the continuous maps }\}\)
\(Int (J)\):
//
Statements:
(
\(\forall j \in Int (J) (r \le f (j))\)
\(\implies\)
\(\forall j \in J (r \le f (j))\)
)
\(\land\)
(
\(\forall j \in Int (J) (f (j) \le r)\)
\(\implies\)
\(\forall j \in J (f (j) \le r)\)
)
//
2: Note
Any non-1-point interval with any closed end is \([r_1, r_2)\), \([r_1, \infty)\), \([r_1, r_2]\), \((r_1, r_2]\), or \((- \infty, r_2]\) with \(r_1 \lt r_2\), and the corresponding interior is \((r_1, r_2)\), \((r_1, \infty)\), \((r_1, r_2)\), \((r_1, r_2)\), or \((- \infty, r_2)\): the \(r_1 = r_2\) case is not considered, because \(Int (J)\) will become empty and the conditions will become vacuous.
\(\forall j \in Int (J) (r \lt f (j))\) implies \(\forall j \in Int (J) (r \le f (j))\), so, it guarantees \(\forall j \in J (r \le f (j))\), but it does not guarantee \(\forall j \in J (r \lt f (j))\): for example, for \(f: [0, 1) \to \mathbb{R}, r \mapsto r\), \(\forall j \in Int (J) (0 \lt f (j))\), but \(f (0) = 0\).
Likewise, \(\forall j \in Int (J) (f (j) \lt r)\) guarantees \(\forall j \in J (f (j) \le r)\) but does not guarantee \(\forall j \in J (f (j) \lt r)\).
3: Proof
Whole Strategy: Step 1: suppose that \(\forall j \in Int (J) (r \le f (j))\); Step 2: see that \(\forall j \in J (r \le f (j))\); Step 3: suppose that \(\forall j \in Int (J) (f (j) \le r)\); Step 4: see that \(\forall j \in J (f (j) \le r)\).
Step 1:
Let us suppose that \(\forall j \in Int (J) (r \le f (j))\).
Step 2:
Let us suppose that the lower end, \(r_1\), is closed.
Let us suppose that \(f (r_1) \lt r\).
Let us take \(B_{f (r_1), (r - f (r_1)) / 2} \subseteq \mathbb{R}\).
As \(f\) was continuous, there would be an open neighborhood of \(r_1\), \(U_{r_1} \subseteq J\), such that \(f (U_{r_1}) \subseteq B_{f (r_1), (r - f (r_1)) / 2}\).
\(U_{r_1} = U'_{r_1} \cap J\) for an open neighborhood of \(r_1\) on \(\mathbb{R}\), \(U'_{r_1} \subseteq \mathbb{R}\), by the definition of subspace topology.
There would be a \(B_{r_1, \delta} \subseteq \mathbb{R}\) such that \(B_{r_1, \delta} \subseteq U'_{r_1}\), by the definition of Euclidean topology.
Let \(\delta\) be small enough such that \(r_1 + \delta / 2 \in Int (J)\), which would be possible, because \(J\) was a non-1-point interval.
\(B_{r_1, \delta} \cap J \subseteq U'_{r_1} \cap J = U_{r_1}\).
So, \(f (B_{r_1, \delta} \cap J) \subseteq B_{f (r_1), (r - f (r_1)) / 2}\).
\(r_1 + \delta / 2 \in B_{r_1, \delta} \cap J\), and \(f (r_1 + \delta / 2) \in B_{f (r_1), (r - f (r_1)) / 2}\).
So, \(f (r_1 + \delta / 2) \lt f (r_1) + (r - f (r_1)) / 2 = (f (r_1) + r) / 2\).
But as \(f (r_1) \lt r\), \(f (r_1) + r \lt 2 r\), and \((f (r_1) + r) / 2 \lt r\).
So, \(f (r_1 + \delta / 2) \lt r\), a contradiction against \(\forall j \in Int (J) (r \le f (j))\).
So, \(r \le f (r_1)\).
Let us suppose that the upper end, \(r_2\), is closed.
Let us suppose that \(f (r_2) \lt r\).
Let us take \(B_{f (r_2), (r - f (r_2)) / 2} \subseteq \mathbb{R}\).
As \(f\) was continuous, there would be an open neighborhood of \(r_2\), \(U_{r_2} \subseteq J\), such that \(f (U_{r_2}) \subseteq B_{f (r_2), (r - f (r_2)) / 2}\).
\(U_{r_2} = U'_{r_2} \cap J\) for an open neighborhood of \(r_2\) on \(\mathbb{R}\), \(U'_{r_2} \subseteq \mathbb{R}\), by the definition of subspace topology.
There would be a \(B_{r_2, \delta} \subseteq \mathbb{R}\) such that \(B_{r_2, \delta} \subseteq U'_{r_2}\), by the definition of Euclidean topology.
Let \(\delta\) be small enough such that \(r_2 - \delta / 2 \in Int (J)\), which would be possible, because \(J\) was a non-1-point interval.
\(B_{r_2, \delta} \cap J \subseteq U'_{r_2} \cap J = U_{r_2}\).
So, \(f (B_{r_2, \delta} \cap J) \subseteq B_{f (r_2), (r - f (r_2)) / 2}\).
\(r_2 - \delta / 2 \in B_{r_2, \delta} \cap J\), and \(f (r_2 - \delta / 2) \in B_{f (r_2), (r - f (r_2)) / 2}\).
So, \(f (r_2 - \delta / 2) \lt f (r_2) + (r - f (r_2)) / 2 = (f (r_2) + r) / 2\).
But as \(f (r_2) \lt r\), \(f (r_2) + r \lt 2 r\), and \((f (r_2) + r) / 2 \lt r\).
So, \(f (r_2 - \delta / 2) \lt r\), a contradiction against \(\forall j \in Int (J) (r \le f (j))\).
So, \(r \le f (r_2)\).
So, as \(J \setminus Int (j)\) contains only \(r_1\) or \(r_2\), \(\forall j \in J (r \le f (j))\).
Step 3:
Let us suppose that \(\forall j \in Int (J) (f (j) \le r)\).
Step 4:
Let us suppose that the lower end, \(r_1\), is closed.
Let us suppose that \(r \lt f (r_1)\).
Let us take \(B_{f (r_1), (f (r_1) - r) / 2} \subseteq \mathbb{R}\).
As \(f\) was continuous, there would be an open ball around \(r_1\), \(B_{r_1, \delta} \subseteq \mathbb{R}\), such that \(f (B_{r_1, \delta} \cap J) \subseteq B_{f (r_1), (f (r_1) - r) / 2}\) and \(r_1 + \delta / 2 \in Int (J)\), as before.
\(r_1 + \delta / 2 \in B_{r_1, \delta} \cap J\), and \(f (r_1 + \delta / 2) \in B_{f (r_1), (f (r_1) - r) / 2}\).
So, \((f (r_1) + r) / 2 = f (r_1) - (f (r_1) - r) / 2 \lt f (r_1 + \delta / 2)\).
But as \(r \lt f (r_1)\), \(2 r \lt f (r_1) + r\), and \(r \lt (f (r_1) + r) / 2\).
So, \(r \lt f (r_1 + \delta / 2)\), a contradiction against \(\forall j \in Int (J) (f (j) \le r)\).
So, \(f (r_1) \le r\).
Let us suppose that the upper end, \(r_2\), is closed.
Let us suppose that \(r \lt f (r_2)\).
Let us take \(B_{f (r_2), (f (r_2) - r) / 2} \subseteq \mathbb{R}\).
As \(f\) was continuous, there would be an open ball around \(r_2\), \(B_{r_2, \delta} \subseteq \mathbb{R}\), such that \(f (B_{r_2, \delta} \cap J) \subseteq B_{f (r_2), (f (r_2) - r) / 2}\) and \(r_2 - \delta / 2 \in Int (J)\), as before.
\(r_2 - \delta / 2 \in B_{r_2, \delta} \cap J\), and \(f (r_2 - \delta / 2) \in B_{f (r_2), (f (r_2) - r) / 2}\).
So, \((f (r_2) + r) / 2 = f (r_2) - (f (r_2) - r) / 2 \lt f (r_2 - \delta / 2)\).
But as \(r \lt f (r_2)\), \(2 r \lt f (r_2) + r\), and \(r \lt (f (r_2) + r) / 2\).
So, \(r \lt f (r_2 - \delta / 2)\), a contradiction against \(\forall j \in Int (J) (f (j) \le r)\).
So, \(f (r_2) \le r\).
So, as \(J \setminus Int (j)\) contains only \(r_1\) or \(r_2\), \(\forall j \in J (f (j) \le r)\).
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 metric is continuous
Topics
About:
metric 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 metric is continuous.
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:
\(M\): \(\in \{\text{ the metric spaces }\}\), with any metric, \(dist\)
//
Statements:
\(dist \in \{\text{ the continuous, metric spaces maps }\}\)
//
2: Note
Of course, this proposition should be able to be proved directly as a metric spaces map, but as we already have the continuousness as the induced topological spaces map, we use it.
3: Proof
Whole Strategy: Step 1: see that \(dist\) is continuous as the induced topological spaces map; Step 2: conclude the proposition.
Step 1:
\(dist: M \times M \to \mathbb{R}\) is continuous with \(M \times M\) regarded as the topological space with the topology as the product of the topology for \(M\) induced by \(dist\), by the proposition that any metric is continuous with respect to the topology induced by the metric.
Step 2:
The product topology is the topology induced by the product metric, by the proposition that for any finite-product metric space, the topology induced by the product metric is the product topology of the topologies induced by the constituent metrics.
The Euclidean topological space, \(\mathbb{R}\), has the topology induced by the Euclidean metric, by definition.
So, \(dist\) as the metric spaces map is continuous, by the proposition that for any metric spaces map, the map is continuous at any point if and only if the map is continuous at the point between the induced topological spaces.
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 finite-product metric space, topology induced by product metric is product topology of topologies induced by constituent metrics
Topics
About:
metric space
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 finite-product metric space, the topology induced by the product metric is the product topology of the topologies induced by the constituent metrics.
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:
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{M_j \vert j \in J\}\): \(M_j \in \{\text{ the metric spaces }\}\) with any metric, \(dist_j\)
\(\times_{j \in J} M_j\): \(= \text{ the finite-product metric space }\) with the finite-product metric, \(dist\)
\(O'\): \(= \text{ the topology induced by } dist\)
\(O\): \(= \text{ the product topology of the topologies induced by } dist_j \text{ s }, \{O_j \vert j \in J\}\)
//
Statements:
\(O' = O\)
//
2: Proof
Whole Strategy: Step 1: take any \(U' \in O'\), and see that \(U' \in O\); Step 2: take any \(U \in O\), and see that \(U \in O'\).
Step 1:
Let \(U' \in O'\) be any.
Let \(u' \in U'\) be any.
There is an open ball, \(B_{u', \epsilon} \subseteq \times_{j \in J} M_j\), such that \(B_{u', \epsilon} \subseteq U'\), by the definition of topology induced by metric.
For \(\beta := \epsilon / \sqrt{\vert J \vert}\), \(\times_{j \in J} B_{u'^j, \beta} \subseteq B_{u', \epsilon}\), by the proposition that for any finite-product metric space and any open ball, there is the product of some open balls contained in the open ball in which (the product) an open ball is contained.
But each \(B_{u'^j, \beta} \subseteq M_j\) is an open neighborhood of \(u'^j\) in the topology for \(M_j\) induced by \(dist_j\), and \(\times_{j \in J} B_{u'^j, \beta} \subseteq \times_{j \in J} M_j\) is an open neighborhood of \(u'\) in the product topology.
As \(\times_{j \in J} B_{u'^j, \beta} \subseteq B_{u', \epsilon} \subseteq U'\), \(U' \in O\), by local criterion for openness.
Step 2:
Let \(U \in O\) be any.
Let \(u \in U\) be any.
\(U = \cup_{j' \in J'} \times_{j \in J} U_{j, j'}\) where \(J'\) is a possibly uncountable index set and \(U_{j, j'} \subseteq M_j\) is open in the topology induced by \(dist_j\): refer to Note for the definition of product topology.
\(u \in \times_{j \in J} U_{j, j'}\) for a \(j' \in J'\).
\(u^j \in U_{j, j'}\), so, there is an open ball, \(B_{u^j, \epsilon_j} \subseteq M_j\), such that \(B_{u^j, \epsilon_j} \subseteq U_{j, j'}\), by the definition of topology induced by metric.
Let \(\epsilon = Min \{\epsilon_j \vert j \in J\}\).
\(B_{u^j, \epsilon} \subseteq U_{j, j'}\) and \(\times_{j \in J} B_{u^j, \epsilon} \subseteq \times_{j \in J} U_{j, j'}\).
\(B_{u, \epsilon} \subseteq \times_{j \in J} B_{u^j, \epsilon}\), by the proposition that for any finite-product metric space and any open ball, there is the product of some open balls contained in the open ball in which (the product) an open ball is contained.
So, \(B_{u, \epsilon} \subseteq \times_{j \in J} B_{u^j, \epsilon} \subseteq \times_{j \in J} U_{j, j'} \subseteq \cup_{j' \in J'} \times_{j \in J} U_{j, j'} = U\).
So, \(U \in O'\), by the definition of topology induced by metric.
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 map from finite-product metric space into metric space continuous at point, induced map with set of components of domain fixed is continuous at point
Topics
About:
metric 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 map from any finite-product metric space into any metric space continuous at any point, the induced map with any set of the components of the domain fixed is continuous at the point.
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:
\(J'\): \(\in \{\text{ the finite index sets }\}\)
\(\{M_j \vert j \in J'\}\): \(M_j \in \{\text{ the metric spaces }\}\) with any metric, \(dist_j\)
\(\times_{j \in J'} M_j\): \(= \text{ the finite-product metric space }\) with the finite-product metric, \(dist\)
\(M\): \(\in \{\text{ the metric spaces }\}\)
\(f'\): \(: \times_{j \in J'} M_j \to M\)
\(J\): \(\subseteq J'\)
\(\pi_1\): \(: \times_{j \in J'} M_j \to \times_{j \in J} M_j\), \(= \text{ the projection }\)
\(\pi_2\): \(: \times_{j \in J'} M_j \to \times_{j \in J' \setminus J} M_j\), \(= \text{ the projection }\)
\(m'\): \(\in \times_{j \in J'} M_j\)
\(g_{m'}\): \(: \times_{j \in J} M_j \to \times_{j \in J'} M_j, m \mapsto g_{m'} (m) \text{ such that } \pi_1 (g_{m'} (m)) = m \land \pi_2 (g_{m'} (m)) = \pi_2 (m')\)
\(f_{m'}\): \(: \times_{j \in J} M_j \to M, m \mapsto f' (g_{m'} (m))\)
//
Statements:
\(f' \in \{\text{ the maps continuous at } m'\}\)
\(\implies\)
\(f_{m'} \in \{\text{ the maps continuous at } \pi_1 (m')\}\)
//
2: Note
When \(f'\) is continuous, \(f_{m'}\) is continuous: for each \(m'' \in \times_{j \in J'} M_j\) such that \(\pi_2 (m') = \pi_2 (m'')\), \(f_{m'} = f_{m''}\) and \(\pi_1 (m'')\) s cover \(\times_{j \in J} M_j\).
3: Proof
Whole Strategy: Step 1: take any \(B_{f_{m'} (\pi_1 (m')), \epsilon}\) and a \(B_{m', \delta}\) such that \(f' (B_{m', \delta}) \subseteq B_{f_{m'} (\pi_1 (m')), \epsilon}\); Step 2: take a \(B_{\pi_1 (m'), \beta}\) such that \(g_{m'} (B_{\pi_1 (m'), \beta}) \subseteq B_{m', \delta}\); Step 3: see that \(f_{m'} (B_{\pi_1 (m'), \beta}) \subseteq B_{f_{m'} (\pi_1 (m')), \epsilon}\).
Step 1:
Let us take any \(B_{f_{m'} (\pi_1 (m')), \epsilon} \subseteq M\).
\(g_{m'} (\pi_1 (m')) = m'\), and \(f_{m'} (\pi_1 (m')) = f' (g_{m'} (\pi_1 (m'))) = f' (m')\).
So, \(B_{f_{m'} (\pi_1 (m')), \epsilon} = B_{f' (m'), \epsilon}\).
As \(f'\) is continuous at \(m'\), there is a \(B_{m', \delta} \subseteq \times_{j \in J'} M_j\) such that \(f' (B_{m', \delta}) \subseteq B_{f_{m'} (\pi_1 (m')), \epsilon}\).
Step 2:
There is \(\beta := \delta / \sqrt{\vert J' \vert}\) such that \(\times_{j \in J'} B_{m'^j, \beta} \subseteq B_{m', \delta}\), by the proposition that for any finite-product metric space and any open ball, there is the product of some open balls contained in the open ball in which (the product) an open ball is contained.
Then, \(\times_{j \in J} B_{m'^j, \beta} \subseteq \times_{j \in J} M_j\) satisfies \(g_{m'} (\times_{j \in J} B_{m'^j, \beta}) \subseteq \times_{j \in J'} B_{m'^j, \beta}\), because for each \(p \in \times_{j \in J} B_{m'^j, \beta}\), \(\pi_1 (g_{m'} (p)) = p\), so, \(\pi_1 (g_{m'} (p))^j = p^j \in B_{m'^j, \beta}\), and \(\pi_2 (g_{m'} (p)) = \pi_2 (m')\), so, \(\pi_2 (g_{m'} (p))^j = \pi_2 (m')^j \in B_{m'^j, \beta}\).
But \(\times_{j \in J} B_{m'^j, \beta} = \times_{j \in J} B_{\pi_1 (m')^j, \beta}\).
\(B_{\pi_1 (m'), \beta} \subseteq \times_{j \in J} B_{\pi_1 (m')^j, \beta}\), by the proposition that for any finite-product metric space and any open ball, there is the product of some open balls contained in the open ball in which (the product) an open ball is contained.
So, \(g_{m'} (B_{\pi_1 (m'), \beta}) \subseteq g_{m'} (\times_{j \in J} B_{\pi_1 (m')^j, \beta}) \subseteq \times_{j \in J'} B_{m'^j, \beta} \subseteq B_{m', \delta}\).
Step 3:
So, \(f_{m'} (B_{\pi_1 (m'), \beta}) = f' \circ g_{m'} (B_{\pi_1 (m'), \beta}) \subseteq f' (B_{m', \delta}) \subseteq B_{f_{m'} (\pi_1 (m')), \epsilon}\).
So, \(f_{m'}\) is continuous at \(\pi_1 (m')\).
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 finite-product metric space and open ball, there is product of open balls contained in open ball in which (product) open ball is contained
Topics
About:
metric 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 finite-product metric space and any open ball, there is the product of some open balls contained in the open ball in which (the product) an open ball is contained.
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:
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{M_j \vert j \in J\}\): \(M_j \in \{\text{ the metric spaces }\}\) with any metric, \(dist_j\)
\(\times_{j \in J} M_j\): \(= \text{ the finite-product metric space }\) with the finite-product metric, \(dist\)
\(m\): \(\in \times_{j \in J} M_j\)
\(B_{m, \epsilon}\): \(\in \{\text{ the open balls around } m\}\)
\(\beta\): \(= \epsilon / \sqrt{\vert J \vert}\)
//
Statements:
\(B_{m, \beta} \subseteq \times_{j \in J} B_{m^j, \beta} \subseteq B_{m, \epsilon}\)
//
2: Proof
Whole Strategy: Step 1: see that \(\times_{j \in J} B_{m^j, \beta} \subseteq B_{m, \epsilon}\); Step 2: see that \(B_{m, \beta} \subseteq \times_{j \in J} B_{m^j, \beta}\).
Step 1:
Let us see that \(\times_{j \in J} B_{m^j, \beta} \subseteq B_{m, \epsilon}\).
Let \(p \in \times_{j \in J} B_{m^j, \beta}\) be any.
\(dist (m, p) = \sqrt{\sum_{j \in J} dist_j (m^j, p^j)^2}\).
As \(p^j \in B_{m^j, \beta}\), \(dist_j (m^j, p^j) \lt \beta\).
So, \(\sqrt{\sum_{j \in J} dist_j (m^j, p^j)^2} \lt \sqrt{\sum_{j \in J} \beta^2} = \sqrt{\sum_{j \in J} \epsilon^2 / \vert J \vert} = \epsilon\).
So, \(p \in B_{m, \epsilon}\).
Step 2:
Let us see that \(B_{m, \beta} \subseteq \times_{j \in J} B_{m^j, \beta}\).
Let \(p \in B_{m, \beta}\) be any.
\(dist (m, p) \lt \beta\).
\(dist (m, p) = \sqrt{\sum_{j \in J} dist_j (m^j, p^j)^2}\).
For each \(j \in J\), \(dist_j (m^j, p^j)^2 \le \sum_{j \in J} dist_j (m^j, p^j)^2\), so, \(dist_j (m^j, p^j) \le \sqrt{\sum_{j \in J} dist_j (m^j, p^j)^2} \lt \beta\).
So, \(p^j \in B_{m^j, \beta}\).
So, \(p \in \times_{j \in J} B_{m^j, \beta}\).
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>
definition of finite-product metric space
Topics
About:
metric space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of finite-product metric space.
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:
\( J\): \(\in \{\text{ the finite index sets }\}\)
\( \{M_j \vert j \in J\}\): \(M_j \in \{\text{ the metric spaces }\}\) with any metric, \(dist_j\)
\( \times_{j \in J} M_j\): \(= \text{ the product set }\)
\( dist\): \(= \text{ the finite-product metric for } \times_{j \in J} M_j\)
\(*(\times_{j \in J} M_j, dist)\)
//
Conditions:
//
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>
definition of finite-product metric
Topics
About:
metric space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of finite-product metric.
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:
\( J\): \(\in \{\text{ the finite index sets }\}\)
\( \{M_j \vert j \in J\}\): \(M_j \in \{\text{ the metric spaces }\}\) with any metric, \(dist_j\)
\( \times_{j \in J} M_j\): \(= \text{ the product set }\)
\(*dist\): \(: \times_{j \in J} M_j \times \times_{j \in J} M_j \to \mathbb{R}, (\times_{j \in J} {m_1}^j, \times_{j \in J} {m_2}^j) \mapsto \sqrt{\sum_{j \in J} dist_j ({m_1}^j, {m_2}^j)^2}\)
//
Conditions:
//
2: Note
Let us see that \(dist\) is indeed a metric.
Let \(m_1, m_2, m_3 \in \times_{j \in J} M_j\) be any.
1) \(0 \le dist (m_1, m_2)\) and \(dist (m_1, m_2) = 0 \iff m_1 = m_2\): \(0 \le \sqrt{\sum_{j \in J} dist_j ({m_1}^j, {m_2}^j)^2} = dist (m_1, m_2)\); if \(dist (m_1, m_2) = 0\), \(dist_j ({m_1}^j, {m_2}^j) = 0\) for each \(j \in J\), so, \({m_1}^j = {m_2}^j\) for each \(j \in J\), so, \(m_1 = m_2\); if \(m_1 = m_2\), \({m_1}^j = {m_2}^j\) for each \(j \in J\), so, \(dist_j ({m_1}^j, {m_2}^j) = 0\) for each \(j \in J\), so, \(dist (m_1, m_2) = \sqrt{\sum_{j \in J} dist_j ({m_1}^j, {m_2}^j)^2} = 0\).
2) \(dist (m_1, m_2) = dist (m_2, m_1)\): \(dist (m_1, m_2) = \sqrt{\sum_{j \in J} dist_j ({m_1}^j, {m_2}^j)^2} = \sqrt{\sum_{j \in J} dist_j ({m_2}^j, {m_1}^j)^2} = dist (m_2, m_1)\).
3) \(dist (m_1, m_3) \le dist (m_1, m_2) + dist (m_2, m_3)\): \(dist (m_1, m_3)^2 = \sum_{j \in J} dist_j ({m_1}^j, {m_3}^j)^2 \le \sum_{j \in J} (dist_j ({m_1}^j, {m_2}^j) + dist_j ({m_2}^j, {m_3}^j))^2 = \sum_{j \in J} (dist_j ({m_1}^j, {m_2}^j)^2 + dist_j ({m_2}^j, {m_3}^j)^2 + 2 dist_j ({m_1}^j, {m_2}^j) dist_j ({m_2}^j, {m_3}^j)) = \sum_{j \in J} dist_j ({m_1}^j, {m_2}^j)^2 + \sum_{j \in J} dist_j ({m_2}^j, {m_3}^j)^2 + 2 \sum_{j \in J} dist_j ({m_1}^j, {m_2}^j) dist_j ({m_2}^j, {m_3}^j) = dist (m_1, m_2)^2 + dist (m_2, m_3)^2 + 2 \sum_{j \in J} dist_j ({m_1}^j, {m_2}^j) dist_j ({m_2}^j, {m_3}^j)\), while \((dist (m_1, m_2) + dist (m_2, m_3))^2 = dist (m_1, m_2)^2 + dist (m_2, m_3)^2 + 2 dist (m_1, m_2) dist (m_2, m_3) = dist (m_1, m_2)^2 + dist (m_2, m_3)^2 + 2 \sqrt{\sum_{j \in J} dist_j ({m_1}^j, {m_2}^j)^2} \sqrt{\sum_{j \in J} dist_j ({m_2}^j, {m_3}^j)^2}\), but \(\sum_{j \in J} dist_j ({m_1}^j, {m_2}^j) dist_j ({m_2}^j, {m_3}^j) \le \sqrt{\sum_{j \in J} dist_j ({m_1}^j, {m_2}^j)^2} \sqrt{\sum_{j \in J} dist_j ({m_2}^j, {m_3}^j)^2}\), by the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space: think of the Euclidean vectors space, \(\mathbb{R}^{\vert J \vert}\), with the Euclidean inner product, and regard \((dist_j ({m_1}^j, {m_2}^j)), (dist_j ({m_2}^j, {m_3}^j)) \in \mathbb{R}^{\vert J \vert}\), then, the left hand side is \(\langle (dist_j ({m_1}^j, {m_2}^j)), (dist_j ({m_2}^j, {m_3}^j) \rangle\) and the right hand side is \(\sqrt{\langle (dist_j ({m_1}^j, {m_2}^j)), (dist_j ({m_1}^j, {m_2}^j)) \rangle} \sqrt{\langle (dist_j ({m_2}^j, {m_3}^j)), (dist_j ({m_2}^j, {m_3}^j)) \rangle}\), so, \(dist (m_1, m_3)^2 \le (dist (m_1, m_2) + dist (m_2, m_3))^2\), so, \(dist (m_1, m_3) \le dist (m_1, m_2) + dist (m_2, m_3)\).
References
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description/proof of that for Euclidean set, taking for \(2\) points, maximum of absolute differences of components of points is metric
Topics
About:
metric 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 Euclidean set, taking for each \(2\) points, the maximum of the absolute differences of the components of the points is a metric.
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:
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(\mathbb{R}^d\): \(= \text{ the Euclidean set }\)
\(f\): \(: \mathbb{R}^d \times \mathbb{R}^d \to \mathbb{R}, (r_1, r_2) \mapsto Max (\{\vert {r_1}^j - {r_2}^j \vert \vert j \in \{1, ..., d\}\})\)
//
Statements:
\(f \in \{\text{ the metrics on } \mathbb{R}^d\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(f\) satisfies the conditions to be a metric.
Step 1:
Let us see that \(f\) satisfies the conditions to be a metric.
Let \(r_1, r_2, r_3 \in \mathbb{R}^d\) be any.
1) \(0 \le f (r_1, r_2)\) and \(f (r_1, r_2) = 0\) if and only if \(r_1 = r_2\): \(0 \le Max (\{\vert {r_1}^j - {r_2}^j \vert \vert j \in \{1, ..., d\}\})\); if \(f (r_1, r_2) = 0\), \({r_1}^j - {r_2}^j = 0\) for each \(j\), which implies that \(r_1 = r_2\); if \(r_1 = r_2\), \({r_1}^j - {r_2}^j = 0\) for each \(j\), which implies that \(f (r_1, r_2) = 0\).
2) \(f (r_1, r_2) = f (r_2, r_1)\): \(f (r_1, r_2) = Max (\{\vert {r_1}^j - {r_2}^j \vert \vert j \in \{1, ..., d\}\}) = Max (\{\vert {r_2}^j - {r_1}^j \vert \vert j \in \{1, ..., d\}\}) = f (r_2, r_1)\).
3) \(f (r_1, r_3) \le f (r_1, s_2) + f (r_2, r_3)\): \(f (r_1, r_3) = Max (\{\vert {r_1}^j - {r_3}^j \vert \vert j \in \{1, ..., d\}\}) = Max (\{\vert {r_1}^j - {r_2}^j + {r_2}^j - {r_3}^j \vert \vert j \in \{1, ..., d\}\}) \le Max (\{\vert {r_1}^j - {r_2}^j \vert + \vert {r_2}^j - {r_3}^j \vert \vert j \in \{1, ..., d\}\}) \le Max (\{\vert {r_1}^j - {r_2}^j \vert \vert j \in \{1, ..., d\}\}) + Max (\{\vert {r_2}^j - {r_3}^j \vert \vert j \in \{1, ..., d\}\})\), by the proposition that for any linearly-ordered ring, any finite number of subsets with any same nonempty finite index set, and the subset as the sum of the subsets with the same index set, the maximum of the subset is equal to or smaller than the sum of the maximums of the subsets and the minimum of the subset is equal to or larger than the sum of the minimums of the subsets, \(= f (r_1, s_2) + f (r_2, r_3)\).
References
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<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for Euclidean set, taking for \(2\) points, sum of absolute differences of components of points is metric
Topics
About:
metric 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 Euclidean set, taking for each \(2\) points, the sum of the absolute differences of the components of the points is a metric.
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:
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(\mathbb{R}^d\): \(= \text{ the Euclidean set }\)
\(f\): \(: \mathbb{R}^d \times \mathbb{R}^d \to \mathbb{R}, (r_1, r_2) \mapsto \sum_{j \in \{1, ..., d\}} \vert {r_1}^j - {r_2}^j \vert\)
//
Statements:
\(f \in \{\text{ the metrics on } \mathbb{R}^d\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(f\) satisfies the conditions to be a metric.
Step 1:
Let us see that \(f\) satisfies the conditions to be a metric.
Let \(r_1, r_2, r_3 \in \mathbb{R}^d\) be any.
1) \(0 \le f (r_1, r_2)\) and \(f (r_1, r_2) = 0\) if and only if \(r_1 = r_2\): \(0 \le \sum_{j \in \{1, ..., d\}} \vert {r_1}^j - {r_2}^j \vert\); if \(f (r_1, r_2) = 0\), \({r_1}^j - {r_2}^j = 0\) for each \(j\), which implies that \(r_1 = r_2\); if \(r_1 = r_2\), \({r_1}^j - {r_2}^j = 0\) for each \(j\), which implies that \(f (r_1, r_2) = 0\).
2) \(f (r_1, r_2) = f (r_2, r_1)\): \(f (r_1, r_2) = \sum_{j \in \{1, ..., d\}} \vert {r_1}^j - {r_2}^j \vert = \sum_{j \in \{1, ..., d\}} \vert {r_2}^j - {r_1}^j \vert = f (r_2, r_1)\).
3) \(f (r_1, r_3) \le f (r_1, s_2) + f (r_2, r_3)\): \(f (r_1, r_3) = \sum_{j \in \{1, ..., d\}} \vert {r_1}^j - {r_3}^j \vert = \sum_{j \in \{1, ..., d\}} \vert {r_1}^j - {r_2}^j + {r_2}^j - {r_3}^j \vert \le \sum_{j \in \{1, ..., d\}} \vert {r_1}^j - {r_2}^j \vert + \vert {r_2}^j - {r_3}^j \vert = \sum_{j \in \{1, ..., d\}} \vert {r_1}^j - {r_2}^j \vert + \sum_{j \in \{1, ..., d\}} \vert {r_2}^j - {r_3}^j \vert = f (r_1, s_2) + f (r_2, r_3)\).
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>
definition of Euclidean set
Topics
About:
set
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of Euclidean set.
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:
\( d\): \(\in \mathbb{N} \setminus \{0\}\)
\(*\mathbb{R}^d\):
//
Conditions:
//
2: Note
The point is that \(\mathbb{R}^d\) is a pure set without any vectors space structure, any metric, any topology, or anything assumed.
Sometimes, it is important to be free from some additional structures usually presupposed onto \(\mathbb{R}^d\).
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 continuous map between metric spaces, restriction of map on compact domain is uniformly continuous
Topics
About:
metric 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 continuous map between any metric spaces with the domain with the induced topology, the restriction of the map on any compact domain is uniformly continuous.
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:
\(M_1\): \(\in \{\text{ the metric spaces }\}\), with the induced topology
\(M_2\): \(\in \{\text{ the metric spaces }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the continuous maps }\}\)
\(K\): \(\in \{\text{ the compact subsets of } M_1\}\), as the metric subspace
//
Statements:
\(f \vert_K: K \to M_2 \in \{\text{ the uniformly continuous maps }\}\)
//
2: Proof
Whole Strategy: Step 1: take any \(\epsilon\) and for each \(k \in K\), \(\delta (k)\) such that \(f (B_{k, \delta (k)}) \subseteq B_{f (k), \epsilon / 2}\), take any finite subcover of \(K\), \(\{B_{k_j, \delta (k_j) / 2}\}\), and take \(\delta := Min (\{\delta (k_j) / 2\})\); Step 2: see that for each \(k \in K\), \(f (B_{k, \delta}) \subseteq B_{f (k), \epsilon}\); Step 3: conclude that proposition.
Step 1:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
For each \(k \in K\), there is a \(\delta (k) \in \mathbb{R}\) such that \(0 \lt \delta (k)\) and \(f (B_{k, \delta (k)}) \subseteq B_{f (k), \epsilon / 2}\), because \(f\) is continuous.
\(\{B_{k, \delta (k) / 2} \vert k \in K\}\) is a open cover of \(K\).
As \(K\) is a compact subset, there is a finite subcover, \(\{B_{k_j, \delta (k_j) / 2} \vert j \in J, k_j \in K\}\), where \(J\) is a finite index set.
Let \(\delta := Min (\{\delta (k_j) / 2 \vert j \in J\})\).
Step 2:
Let \(k \in K\) be any.
Let us see that \(f (B_{k, \delta}) \subseteq B_{f (k), \epsilon}\).
\(k \in B_{k_j, \delta (k_j) / 2}\) for a \(j \in J\).
As \(f (B_{k_j, \delta (k_j)}) \subseteq B_{f (k_j), \epsilon / 2}\), \(dist (f (k_j), f (k)) \lt \epsilon / 2\).
Let \(k' \in B_{k, \delta}\) be any.
\(dist (k', k_j) \le dist (k', k) + dist (k, k_j) \lt \delta + \delta (k_j) / 2 \le \delta (k_j) / 2 + \delta (k_j) / 2 = \delta (k_j)\), which means that \(k' \in B_{k_j, \delta (k_j)}\).
So, as \(f (B_{k_j, \delta (k_j)}) \subseteq B_{f (k_j), \epsilon / 2}\), \(dist (f (k'), f (k_j)) \lt \epsilon / 2\).
\(dist (f (k'), f (k)) \le dist (f (k'), f (k_j)) + dist (f (k_j), f (k)) \lt \epsilon / 2 + \epsilon / 2 = \epsilon\).
That means that \(f (B_{k, \delta}) \subseteq B_{f (k), \epsilon}\).
Step 3:
So, \(f \vert_K (B_{k, \delta} \cap K) \subseteq B_{f (k), \epsilon}\).
But \(B_{k, \delta} \cap K\) is \(B_{k, \delta}\) on \(K\), by the subspace metric.
So, \(f \vert_K (B_{k, \delta}) \subseteq B_{f (k), \epsilon}\), where \(B_{k, \delta}\) here is the open ball on \(K\).
So, \(f \vert_K\) is uniformly continuous.
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>
definition of uniformly continuous map between metric spaces
Topics
About:
metric space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of uniformly continuous map between metric spaces.
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:
\( M_1\): \(\in \{\text{ the metric spaces }\}\)
\( M_2\): \(\in \{\text{ the metric spaces }\}\)
\(*f\): \(: M_1 \to M_2\)
//
Conditions:
\(\forall \epsilon \in \mathbb{R} \text{ such that } 0 \lt \epsilon (\exists \delta \in \mathbb{R} \text{ such that } 0 \lt \delta (\forall m_1 \in M_1 (f (B_{m_1, \delta}) \subseteq B_{f (m_1), \epsilon})))\)
//
2: Note
The point is that \(\delta\) does not depend on \(m_1\).
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 metric spaces map, map is continuous at point iff map is continuous at point between induced topological spaces
Topics
About:
metric 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 metric spaces map, the map is continuous at any point if and only if the map is continuous at the point between the induced topological spaces.
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:
\(M_1\): \(\in \{\text{ the metric spaces }\}\)
\(M_2\): \(\in \{\text{ the metric spaces }\}\)
\(T_1\): \(= \text{ the topological space induced by } M_1\)
\(T_2\): \(= \text{ the topological space induced by } M_2\)
\(m\): \(\in M_1\)
\(f\): \(: M_1 \to M_2\)
\(f'\): \(: T_1 \to T_2, t \mapsto f (t)\)
//
Statements:
\(f \in \{\text{ the maps continuous at } m\}\)
\(\iff\)
\(f' \in \{\text{ the maps continuous at } m\}\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(f\) is continuous at \(m\); Step 2: see that \(f'\) is continuous at \(m\); Step 3: suppose that \(f'\) is continuous at \(m\); Step 4: see that \(f\) is continuous at \(m\).
Step 1:
Let us suppose that \(f\) is continuous at \(m\).
Step 2:
Let \(U_{f (m)} \subseteq T_2\) be any open neighborhood of \(f (m)\).
There is an \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\) and \(B_{f (m), \epsilon} \subseteq U_{f (m)}\), by the definition of topology induced by metric.
There is a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and \(f (B_{m, \delta}) \subseteq B_{f (m), \epsilon}\), because \(f\) is continuous.
So, \(f' (B_{m, \delta}) = f (B_{m, \delta}) \subseteq B_{f (m), \epsilon} \subseteq U_{f (m)}\).
But \(B_{m, \delta} \subseteq T_1\) is an open neighborhood of \(m\), by Note for the definition of topology induced by metric.
So, \(f'\) is continuous at \(m\).
Step 3:
Let us suppose that \(f'\) is continuous at \(m\).
Step 4:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
\(B_{f (m), \epsilon} \subseteq T_2\) is open, by Note for the definition of topology induced by metric.
There is an open neighborhood of \(m\), \(U_m \subseteq T_1\), such that \(f' (U_m) \subseteq B_{f (m), \epsilon}\), because \(f'\) is continuous.
There is a \(\delta \in \mathbb{R}\) such that \(0 \lt \delta\) and \(B_{m, \delta} \subseteq U_m\), by the definition of topology induced by metric.
So, \(f (B_{m, \delta}) \subseteq f (U_m) = f' (U_m) \subseteq B_{f (m), \epsilon}\).
So, \(f\) is continuous at \(m\).
References
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