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# 4.2: Open Sets

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## Definition

We say a set $$U \subset \mathbb{R}$$ is open if for every $$x \in U$$ there exists $$\epsilon>0$$ such that

$(x-\epsilon, x+\epsilon) \subset U.$

## Proposition $$\PageIndex{1}$$

Every open interval $$I$$ is an open set.

Proof

Suppose $$I=(a, b),$$ where $$a<b$$ are extended real numbers. Given $$x \in I,$$ let $$\epsilon$$ be the smaller of $$x-a$$ and $$b-x .$$ Suppose $$y \in(x-\epsilon, x+\epsilon) .$$ If $$b=+\infty,$$ then $$b>y ;$$ otherwise, we have

$b-y>b-(x+\epsilon)=(b-x)-\epsilon \geq(b-x)-(b-x)=0,$

so $$b>y .$$ If $$a=-\infty,$$ then $$a<y ;$$ otherwise,

$y-a>(x-\epsilon)-a=(x-a)-\epsilon \geq(x-a)-(x-a)=0,$

so $$a<y .$$ Thus $$y \in I$$ and $$I$$ is an open set. $$\quad$$ Q.E.D.

Note that $$\mathbb{R} \text { is an open set (it is, in fact, the open interval }(-\infty,+\infty)),$$ as is $$\emptyset$$ (it satisfies the definition trivially).

## Proposition $$\PageIndex{2}$$

Suppose $$A$$ is a set and, for each $$\alpha \in A, U_{\alpha}$$ is an open set. Then

$\bigcup_{\alpha \in A} U_{\alpha}$

is an open set.

Proof

Let $$x \in \cup_{\alpha \in A} U_{\alpha} .$$ Then $$x \in U_{\alpha}$$ for some $$\alpha \in A .$$ Since $$U_{\alpha}$$ is open, there exists an $$\epsilon>0$$ such that $$(x-\epsilon, x+\epsilon) \subset U_{\alpha} .$$ Thus

$(x-\epsilon, x+\epsilon) \subset U_{\alpha} \subset \bigcup_{\alpha \in A} U_{\alpha}.$

Hence $$\bigcup_{\alpha \in A} U_{\alpha}$$ is open. $$\quad$$ Q.E.D.

## Proposition $$\PageIndex{3}$$

Suppose $$U_{1}, U_{2}, \ldots, U_{n}$$ is a finite collection of open sets. Then

$\bigcap_{i=1}^{n} U_{i}$

is open.

Proof

Let $$x \in \bigcap_{i=1}^{n} U_{i} .$$ Then $$x \in U_{i}$$ for every $$i=1,2, \ldots, n .$$ For each $$i$$, choose $$\epsilon_{i}>0$$ such that $$\left(x-\epsilon_{i}, x+\epsilon_{i}\right) \subset U_{i} .$$ Let $$\epsilon$$ be the smallest of $$\epsilon_{1}, \epsilon_{2}, \ldots, \epsilon_{n} .$$ Then $$\epsilon>0$$ and

$(x-\epsilon, x+\epsilon) \subset\left(x-\epsilon_{i}, x+\epsilon_{i}\right) \subset U_{i}$

for every $$i=1,2, \ldots, n .$$ Thus

$(x-\epsilon, x+\epsilon) \subset \bigcap_{i=1}^{n} U_{i}.$

Hence $$\bigcap_{i=1}^{n} U_{i}$$ is an open set. $$\quad$$ Q.E.D.

## Definition

Let $$A \subset \mathbb{R} .$$ We say $$x \in A$$ is an interior point of $$A$$ if there exists an $$\epsilon>0$$ such that $$(x-\epsilon, x+\epsilon) \subset A .$$ We call the set of all interior points of $$A$$ the interior of $$A,$$ denoted $$A^{\circ} .$$

## Exercise $$\PageIndex{1}$$

Show that if $$A \subset \mathbb{R},$$ then $$A^{\circ}$$ is open.

## Exercise $$\PageIndex{2}$$

Show that $$A$$ is open if and only if $$A=A^{\circ}$$.

## Exercise $$\PageIndex{3}$$

Let $$U \subset \mathbb{R}$$ be a nonempty open set. Show that sup $$U \notin U$$ and $$\inf U \notin U$$.

This page titled 4.2: Open Sets is shared under a CC BY-NC-SA 1.0 license and was authored, remixed, and/or curated by Dan Sloughter via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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