Proof Verification: Finding A Ball Strictly Contained In An Open Set Of A Metric Space

Question

Problem: Let $X$ be a metric space and let $A$ be an open set of $X$ containing a point $x \in X$. Prove that there exists an $\epsilon > 0$ such that $B_{\epsilon}(x)$ is strictly contained in $A$.

Proof Attempt:

Case 1: $\partial A = \emptyset$

Since $X$ is a metric space, this implies that $A$ is clopen. The only clopen sets of a metric space are $\emptyset$ and the entire space. $A$ contains $x$, so it cannot be empty and thus $A = X$, so any $\epsilon > 0$ will suffice.

Case 2: $\partial A \neq \emptyset$

Let $\displaystyle \epsilon = \frac{1}{2}\inf_{p \in \partial A}{d(p,x)}$, where $d$ is the metric of $X$. Note that $\epsilon \neq 0$ or else this would imply that $x \in \partial A$, which contradicts the hypothesis that $A$ contains $x$ and that $A$ is an open set. So $\epsilon > 0$. Then $B_{\epsilon}(x)$ is strictly contained in $A$ (I'm not sure how to justify this part). $\blacksquare$

Is this proof correct? How do I finish the proof? Thanks.

Answer 1

Your proof is flawed. The part that says "the only clopen sets of a metric space are $\emptyset$ and the entire space" is true only when $X$ is connected.

Moreover, your statement works only if $|A| \geq 2$.

Here's a revised argument:

By the definition of a base for the metric space $X$, you can find an open ball $x \in B_{\rho}(x^*) \subseteq A$. Since $x$ is an internal point, you can assume, W.L.O.G., that $\exists \epsilon >0:B_{\epsilon}(x) \subseteq A$.

If $B_{\epsilon}(x) = A$, since a metric space is Hausdorff and $A$ has at least two points, let's say $x,x' \in A$, you can find two open sets $x\in U$ and $x' \in V$ separating them from each other. Now that $U \neq A$, find a ball $B_{\delta}(x) \subseteq U \neq A$ and you're done.

If $A$ has only one point, your statement is wrong as cleverly noted by Adayah. Indeed, if $A=\{x\}$ it's obvious that it cannot strictly contain an open ball.

Answer 2

What's your definition of open?

Mine is a set, $A$, open if every point is an interior point. And my definition of interior point, $x$, is if there is an open ball with $x$ at its center that is completely contained in $A$.

Which makes this statement true by definition.

By your given proof it seems you are using a definition: $A$ is open if it has no boundary points and a boundary point, I am assuming, is defined as a point $x \in X$ so that every open ball of $x$ contains points that are in $A$ and points that are not in $A$.

Okay... so assume $A$ is open and $x\in A$. Then $x$ is not a boundary point because no boundary points of $A$ exist. Now every open ball centered on $x$ contain $x \in A$ so every open ball centered on $x$ contains a point in $A$. And as $x$ is not a boundary point:

• It is not true that every open ball centered at $x$ will contain points in $A$ and will contain points not in $A$.
• Therefore there will exist an open ball, $B_\epsilon(x)$, for which it is not true that will contain both points in $A$ and points not in $A$.
• Therefore for $B_\epsilon(x)$ it will be true that either all points are in $A$ or that all points are not in $A$.
• However $x \in A$ and $x \in B_\epsilon(x)$ so it is not true that all points are not in $A$.
• So it is true that all points in $B_\epsilon(x)$ are in $A$.

So $B_\epsilon(x)$ is completely contained in $A$.