# Prove: A subset V of $\mathbb{R}$ is open iff V is equal to a union of open intervals

The proof of the theorem is given to me in the book but I need some clarification about specific aspects of the proof that the book thinks is trivial:

$\Rightarrow$ Assume V is a open set of $\mathbb{R}$

If V is the empty set then V is trivially the union of an empty collection of open intervals

If V is nonempty then for each $x\in V$ there's an open interval $I_{x}$ such that $x\in I_{x} \subseteq V$

It is easily seen that $V=\cup \{I_{x} : x\in V\}$

My question is how is that easily seen? Why does that show that V is equal to a union of open intervals?

$\Leftarrow$ Assume there's a collection of open intervals $\{I_{\alpha}: \alpha\in \Lambda\}$ such that $V=\cup \{I_{\alpha}: \alpha \in \Lambda \}$

Let $x\in V$

Then there's some $\beta \in V$ for which $x\in I_{\beta}$

Clearly $I_{\beta} \subseteq V$. Hence V is an open subset of $\mathbb{R}$

Why is $I_{\beta}$ a subset of V?

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In other words, we have the containments $\bigcup_{x \in V} I_x \subseteq V$ and $V \subseteq \bigcup_{x \in V} I_x$.

The following principle is used over and over again, and is extremely useful:

• A containment $\bigcup_{a \in I} S_a \subseteq T$ holds if and only if $S_a \subseteq T$ for all $a \in I$.

Use this whenever you're trying to show that a union is contained in something else. In the present case, it means $\bigcup_{x \in V} I_x \subseteq V$ is established as soon as $I_x \subseteq V$ for every $x \in V$. Which we have.

The other containment is easy. To show $V \subseteq \bigcup_{x \in V} I_x$, we have to show that every $y \in V$ belongs to the union. But for such $y$ we have $y \in I_y$, and since $I_y \subseteq \bigcup_{x \in V} I_x$, we have $y \in \bigcup_{x \in V} I_x$.

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An attempt at answering with few symbols:

Recall that a set of real numbers $V$ is open provided that, for any point in $V$, you can find an open interval that both contains the point and is entirely contained in $V$.

The forward direction of your proposition:

Let $V$ be a non-empty open set. This means that for each point in $V$, we can find a corresponding open interval that both contains the point and is contained in $V$. Do this for each point in $V$, and then take the union of all these open intervals, which we claim is precisely $V$.

Note that this union of open intervals is contained in $V$: after all, each of the individual intervals is contained in $V$, so their union is as well; note further that the union contains all of $V$, since for each point in $V$, we included in our union an open interval containing that point.

These two containments combine to give the desired result.

The reverse direction of your proposition:

Suppose $V$ is a non-empty union of open intervals. Pick any point in $V$; by the definition of our set (i.e., as a non-empty union of open intervals), there must be some open interval containing this point. Since the point is now in that open interval, we have found an open set (namely, that particular open interval) contained in $V$, which contains the chosen point. Since this can be done for any point in $V$, we have proven that $V$ is an open set (in accordance with the first paragraph of my response here).

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