4.2: Open Sets

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 \(\inf U \notin U\).