798: For Set and 2 Topologies, iff There Is Common Open Cover and Each Open Subset of Each Element of Cover in One Topology Is Open in the Other and Vice Versa, Topologies Are Same

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description/proof of that for set and 2 topologies, iff there is common open cover and each open subset of each element of cover in one topology is open in the other and vice versa, topologies are same

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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 topological space.
• The reader knows a definition of subspace topology of subset of topological space.
• The reader admits the proposition that for any topological space and any open cover, any subset is open iff the intersection of the subset and each element of the open cover is open.
• The reader admits 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.

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Target Context

• The reader will have a description and a proof of the proposition that for any set and any 2 topologies for the set, iff there is a common open cover and each open subset of each element of the cover in one topology is open in the other topology and vice versa, the topologies are the same.

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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:
$S$: $\in \{\text{ the sets }\}$
$O_1$: $\in \{\text{ the topologies for }S\}$
$O_2$: $\in \{\text{ the topologies for }S\}$
//

Statements:
( $\mathrm{\exists }\{U_{\alpha }|A\}\in \{\text{ the open covers in }{O}_1\}\cap \{\text{ the open covers in }{O}_2\}$
$\wedge$
$\mathrm{\forall }\alpha \in A$
(
$\mathrm{\forall }U\in \{\text{ the open subsets of }{U}_{\alpha }\text{ in }{O}_1\}(U\in \{\text{ the open subsets of }{U}_{\alpha }\text{ in }{O}_2\})$
$\wedge$
$\mathrm{\forall }U\in \{\text{ the open subsets of }{U}_{\alpha }\text{ in }{O}_2\}(U\in \{\text{ the open subsets of }{U}_{\alpha }\text{ in }{O}_1\})$
)
)
$⟺$
$O_1=O_2$
//

"open subset of $U_{\alpha }$ in $O_j$" strictly means that it is an open subset of the topological subspace $U_{\alpha }$ induced by $O_j$, but as $U_{\alpha }$ is open in $O_j$, in fact, it is also open in $O_j$.

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2: Note

This proposition says "Sameness of topologies can be checked locally." so to speak.

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3: Proof

Whole Strategy: Step 1: suppose that there is a common open cover and each open subset of each element of the cover in one topology is open in the other topology, take any open subset, $U_1\subseteq S$, in $O_1$, see that each $U_1\cap U_{\alpha }$ is open in $O_2$, and conclude that $U_1$ is open in $O_2$; Step 2: likewise, conclude that any open subset, $U_2\subseteq S$, in $O_2$ is open in $O_1$; Step 3: conclude that $O_1=O_2$; Step 4: suppose that $O_1=O_2$, and conclude that there is a common open cover and each open subset of each element of the cover in one topology is open in the other topology.

Step 1:

Let us suppose that there is a common open cover and each open subset of each element of the cover in one topology is open in the other topology.

Let $U_1\subseteq S$ be any open subset in $O_1$.

We are going to see that $U_1$ is open in $O_2$.

By the proposition that for any topological space and any open cover, any subset is open iff the intersection of the subset and each element of the open cover is open, what we need to do is to see that each $U_1\cap U_{\alpha }$ is open in $O_2$.

As $U_1\cap U_{\alpha }\subseteq U_{\alpha }$ is open in the subspace topology of $U_{\alpha }$ induced by $O_1$, $U_1\cap U_{\alpha }\subseteq U_{\alpha }$ is open in the subspace topology induced by $O_2$, by the supposition. As $U_{\alpha }\subseteq S$ is open in $O_2$, $U_1\cap U_{\alpha }\subseteq S$ is open in $O_2$, 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.

So, $U_1$ is indeed open in $O_2$.

Step 2:

Likewise, any open subset, $U_2\subseteq S$, in $O_2$ is open in $O_1$, by the symmetry.

Step 3:

Step 1 and Step 2 mean that $O_1=O_2$.

Step 4:

Let us suppose that $O_1=O_2$.

Of course, there is a common open cover.

Of course, each open subset of each element of the cover in one topology is open in the other topology.

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References

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