848: For Maps Between Arbitrary Subspaces of Topological Spaces Continuous at Corresponding Points, Composition Is Continuous at Point

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description/proof of that for maps between arbitrary subspaces of topological spaces continuous at corresponding points, composition is continuous at point

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of map continuous at point.
• The reader knows a definition of subspace topology of subset of topological space.

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Target Context

• The reader will have a description and a proof of the proposition that for any maps between any arbitrary subspaces of any topological spaces continuous at any corresponding points, the composition is continuous at the point.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$T_3$: $\in \{\text{ the topological spaces }\}$
$S_1$: $\in \{\text{ the topological subspaces of }{T}_1\}$
$S_2$: $\in \{\text{ the topological subspaces of }{T}_2\}$
$S_2^{\prime }$: $\in \{\text{ the topological subspaces of }{T}_2\}$, such that $S_2\subseteq S_2^{\prime }$
$S_3$: $\in \{\text{ the topological subspaces of }{T}_3\}$
$p$: $\in S_1$
$f_1$: $:S_1\to S_2$, $\in \{\text{ the maps continuous at }p\}$
$f_2$: $:S_2^{\prime }\to S_3$, $\in \{\text{ the maps continuous at }{f}_1(p)\}$
$f_2\circ f_1$: $:S_1\to S_3$
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Statements:
$f_2\circ f_1\in \{\text{ the maps continuous at }p\}$
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2: Note

It is crucial that $S_2$ and $S_2^{\prime }$ are regarded to be some subspaces of the same $T_2$: if $f_2$ is continuous with $S_2^{\prime }$ regarded as a topological subspace of another topological space, $T_2^{\prime }$, with the same set but with a different topology, this proposition cannot be applied.

For an obvious example, if we were allowed to choose a different topology for $T_2^{\prime }$, I would choose the discrete topology for $T_2^{\prime }$, which would make any $f_2$ be continuous (any map from any discrete topological space is continuous), and the proposition would imply that for whatever $f_2$, $f_2\circ f_1$ would be continuous, which is of course not true.

We will sometimes call a composition of continuous maps "legitimate chain of continuous maps" when the requirements for this proposition are satisfied (when $S_2$ and $S_2^{\prime }$ are regarded to be some topological subspaces of the same $T_2$ together with $S_2\subseteq S_2^{\prime }$). That is because we (or at least I) tend to slip in checking the requirements.

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3: Proof

Whole Strategy: take any open neighborhood of $f_2\circ f_1(p)$, $U_{f_2\circ f_1(p)}\subseteq S_3\subseteq T_3$, and an open neighborhood of $p$, $U_p\subseteq S_1\subseteq T_1$, such that $f_2\circ f_1(U_p)\subseteq U_{f_2\circ f_1(p)}$; Step 1: take any open neighborhood of $f_2\circ f_1(p)$, $U_{f_2\circ f_1(p)}\subseteq S_3\subseteq T_3$; Step 2: take an open neighborhood of $f_1(p)$, $U_{f_1(p)}^{\prime }\subseteq S_2^{\prime }\subseteq T_2$, such that $f_2(U_{f_1(p)}^{\prime })\subseteq U_{f_2\circ f_1(p)}$; Step 3: take $U_{f_1(p)}:=U_{f_1(p)}^{\prime }\cap S_2\subseteq S_2\subseteq T_2$ and see that it is an open neighborhood of $f_1(p)$ on $S_2$; Step 4: take an open neighborhood of $p$, $U_p\subseteq S_1\subseteq T_1$, such that $f_1(U_p)\subseteq U_{f_1(p)}$, and see that $f_2\circ f_1(U_p)\subseteq U_{f_2\circ f_1(p)}$.

Step 1:

Let us take any open neighborhood of $f_2\circ f_1(p)$, $U_{f_2\circ f_1(p)}\subseteq S_3\subseteq T_3$.

Step 2:

As $f_2$ is continuous at $f_1(p)$, there is an open neighborhood of $f_1(p)$, $U_{f_1(p)}^{\prime }\subseteq S_2^{\prime }\subseteq T_2$, such that $f_2(U_{f_1(p)}^{\prime })\subseteq U_{f_2\circ f_1(p)}$.

Step 3:

Let us take $U_{f_1(p)}:=U_{f_1(p)}^{\prime }\cap S_2\subseteq S_2\subseteq T_2$.

Let us see that $U_{f_1(p)}$ is an open neighborhood of $f_1(p)$ on $S_2$.

As $U_{f_1(p)}^{\prime }$ is an open neighborhood of $f_1(p)$ on $S_2^{\prime }$, $U_{f_1(p)}^{\prime }=U^{\prime }\cap S_2^{\prime }$ where $U^{\prime }\subseteq T_2$ is an open subset of $T_2$, by the definition of subspace topology.

$U_{f_1(p)}=U_{f_1(p)}^{\prime }\cap S_2=U^{\prime }\cap S_2^{\prime }\cap S_2=U^{\prime }\cap S_2$, because $S_2\subseteq S_2^{\prime }$.

That means that $U_{f_1(p)}$ is an open subset of $S_2$, by the definition of subspace topology.

$f_1(p)\in U_{f_1(p)}$, because $f_1(p)\in U_{f_1(p)}^{\prime }$ and $f_1(p)\in \cap S_2$.

So, $U_{f_1(p)}$ is an open neighborhood of $f_1(p)$ on $S_2$.

Step 4:

As $f_1$ is continuous at $p$, there is an open neighborhood of $p$, $U_p\subseteq S_1\subseteq T_1$, such that $f_1(U_p)\subseteq U_{f_1(p)}$.

$f_2\circ f_1(U_p)\subseteq f_2(U_{f_1(p)})\subseteq f_2(U_{f_1(p)}^{\prime })\subseteq U_{f_2\circ f_1(p)}$.

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References

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