365: Pair of Open Sets of Connected Topological Space Is Finite-Open-Sets-Sequence-Connected

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A description/proof of that pair of open sets of connected topological space is finite-open-sets-sequence-connected

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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: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of connected topological space.
• The reader knows a definition of finite-open-sets-sequence-connected pair of open sets.

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Target Context

• The reader will have a description and a proof of the proposition that any pair of open sets of any connected topological space is finite-open-sets-sequence-connected.

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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: Description

For any connected topological space, $T$, and any pair of open sets, $U_1,U_2\subseteq T$, $U_1$ and $U_2$ are finite-open-sets-sequence-connected.

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2: Proof

Take any open cover, $S_c=\{U_{\alpha }\}$, of $T$ that includes $U_1$ and $U_2$, which is always possible because we can take the set of neighborhoods of all the point, with $U_1$ and $U_2$ added. Take the equivalence class, $S_e=\{U_{\beta }\}\subseteq S_c$, of the open cover such that each element of the equivalence class is finite-open-sets-sequence-connected with $U_1$ via some elements of $S_c$, which is obviously an equivalence class.

If $S_e$ did not equal $S_c$, $S_r:=S_c\setminus S_e=\{U_{\gamma }\}$ would be nonempty. As $S_e\cup S_r=S_c$ is an open cover, $({\cup }_{\beta }{U}_{\beta })\cup ({\cup }_{\gamma }{U}_{\gamma })=T$, but as $T$ is connected, $({\cup }_{\beta }{U}_{\beta })\cap ({\cup }_{\gamma }{U}_{\gamma })\ne \mathrm{\varnothing }$ because otherwise, $T$ would be the disjoint union of open sets, ${\cup }_{\beta }{U}_{\beta }$ and ${\cup }_{\gamma }{U}_{\gamma }$. So, there would be a point, $p\in T$, such that $p\in {\cup }_{\beta }{U}_{\beta }$ and $p\in {\cup }_{\gamma }{U}_{\gamma }$, which means that $p\in U_{\beta }$ for an $\beta$ and $p\in U_{\gamma }$ for a $\gamma$, so, $p\in U_{\beta }\cap U_{\gamma }$, which is a contradiction, because as $U_{\gamma }$ would share a point with $U_{\beta }$, $U_{\gamma }$ should be finite-open-sets-sequence-connected with $U_1$. So, $S_e$ equals $S_c$.

So, $U_2\in S_e$, and $U_1$ and $U_2$ are finite-open-sets-sequence-connected.

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3: Note

Although not every connected topological space is path-connected, any pair of open sets of any connected topological space is connected by way of open sets.

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References

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