797: For Topological Space and Open Cover, Subset Is Open iff Intersection of Subset and Each Element of Open Cover Is Open

<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 open cover, subset is open iff intersection of subset and each element of open cover is open

---

Topics

About: topological space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of topological space.
• The reader admits the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.

---

Target Context

• The reader will have a description and a proof of 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.

---

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:
$T$: $\in \{\text{ the topological spaces }\}$
$A$: $\in \{\text{ the possibly uncountable index sets }\}$
$\{U_{\alpha }|\alpha \in A\}$: $\in \{\text{ the open covers of }T\}$
$S\subseteq T$:
//

Statements:
$S\in \{\text{ the open subsets of }T\}$
$⟺$
$\mathrm{\forall }\alpha \in A(S\cap U_{\alpha }\in \{\text{ the open subsets of }T\})$
//

---

2: Note

This proposition says "Openness of subset can be checked locally." so to speak.

---

3: Proof

Whole Strategy: Step 1: suppose that $S$ is open and see that $S\cap U_{\alpha }$ is open; Step 2: suppose that $S\cap U_{\alpha }$ is open, see that $S=T\cap S=({\cup }_{\alpha \in A}{U}_{\alpha })\cap S$, and see that $S$ is open.

Step 1:

Let us suppose that $S$ is open on $T$.

For each $\alpha \in A$, $S\cap U_{\alpha }$ is open on $T$ as a finite intersection of open subsets.

Step 2:

Let us suppose that for each $\alpha \in A$, $S\cap U_{\alpha }$ is open on $T$.

$S=T\cap S=({\cup }_{\alpha \in A}{U}_{\alpha })\cap S={\cup }_{\alpha \in A}(U_{\alpha }\cap S)$, by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset, which is open on $T$ as a union of open subsets.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>