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.

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    $\begingroup$ Nope. The only clopen sets of a metric spaces are $\emptyset$ and itself only when it's connected. Plus, why do you need to show that $A$ is clopen when $A$ is open? Just use the fact that open balls are a basis for the metric topology. That's all that there's to it, I think. $\endgroup$ – stressed out Jul 11 '18 at 19:28
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    $\begingroup$ The claim is false if $A$ contains just $x$ (which is possible if $x$ is isolated). $\endgroup$ – Adayah Jul 11 '18 at 19:32
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    $\begingroup$ @Arthur How do you define a topology induced by the metric in a metric space? The only definition that I know is that its the topology generated by open balls as its basis. $\endgroup$ – stressed out Jul 11 '18 at 19:37
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    $\begingroup$ @Arthur I'm not sure I'm following you. When I have a metric on $X$, the metric topology is defined to be the topology generated with open balls as its basis. It's the definition. There's nothing to prove really. Am I missing something? $\endgroup$ – stressed out Jul 11 '18 at 19:40
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    $\begingroup$ @Arthur And how do you define a "metric-open" set when you haven't defined a topology yet? By the way, the definition I used is very standard. It's the standard definition of the topology induced by a metric on a space. $\endgroup$ – stressed out Jul 11 '18 at 19:51

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.

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    $\begingroup$ Doesn't it disturb you a little that $B_{\epsilon}(x)$ should be strictly contained in $A$? $\endgroup$ – Adayah Jul 11 '18 at 19:53
  • $\begingroup$ OK, I see that using the definition of basis, the proof is very trivial, but how do we address Adayah's concern about isolated points? Suppose $X$ is a metric space and $A$ is a subset of $X$ which contains an isolated point $x$. Now give $A$ the subspace topology. First of all, is $A$ a metric space? (I think it should be, because it inherited the topology from $X$). If yes, then considering $A$ as a metric space, $\{x\}$ is an open set of $A$. Now we contradict the claim that we can find a ball around $x$ strictly contained in $A$. So is the claim actually false? $\endgroup$ – Frederic Chopin Jul 11 '18 at 19:57
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    $\begingroup$ 1. $B_{\epsilon}(x)$ is said to be strictly contained in $A$ if $B_{\epsilon}(x) \color{red}{\subsetneq} A$. If you've got that $B_{\epsilon}(x) \subseteq A$ you still have to show that $B_{\epsilon}(x) \neq A$. 2. If $x$ is an isolated point, then $\{ x \}$ is open in $X$ (that is the definition of an isolated point, after all, and not only so in a metric space; it works in an arbitrary topological space). If we take $A = \{ x \}$, then the claim is obviously false, since we cannot have $B_{\epsilon}(x) \subsetneq \{ x \}$. $\endgroup$ – Adayah Jul 11 '18 at 20:02
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Adayah Jul 11 '18 at 20:26
  • $\begingroup$ Even if one assumes that "open" is defined with reference to "base" (which is but one possible definition, and probably not the first one given in most developments, particularly if metric spaces are covered before abstract topological spaces), it is not clear by definition that $A$ has to contain a ball around $x$ specifically. What that definition gives us is just that $x\in B_\varepsilon(y)\subseteq A$ for some $\varepsilon$ and $y$. Showing that we can always choose $y$ to be $x$ requires an additional reasoning step, appealing to the triangle equality. $\endgroup$ – Henning Makholm Jul 11 '18 at 22:38

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 $A$.

  • $\begingroup$ I think you fell into the same trap as me. xD The problem asks that $B_{\epsilon}(x)$ should be strictly contained in $A$, which is of course false if $A$ has only one element. $A$ must at least have two elements. Also, the logical equivalence "$A$ is open if and only if it has no boundary points" is very easy to see. If $a \in A$ and $A$ has no boundary points, then there must exist some ball centered at $a$ not intersecting $A^c$. So, it completely lies inside $A$. Conversely, if $a \in A$ is arbitrary and all balls centered at it are contained in $A$, then $a$ is not a boundary point. $\endgroup$ – stressed out Jul 12 '18 at 1:40
  • $\begingroup$ I also want to add that my above comment was not a criticism of your post (I wanted to say this but I couldn't write a longer comment). I just wanted to explain why all definitions for an open set in a metric space can easily be seen to be the same. $\endgroup$ – stressed out Jul 12 '18 at 1:43
  • $\begingroup$ "strictly" could be false non-connected metric spaces and I'm pretty certain that is not what the original question was asking. " I just wanted to explain why all definitions for an open set in a metric space can easily be seen to be the same. " I thought I went into that in excruciating detail. $\endgroup$ – fleablood Jul 12 '18 at 5:32
  • $\begingroup$ I'm afraid that is what the original question was asking (read it again). I didn't notice it either at first, until Adayah pointed it out to me. Maybe you went into that in excruciating detail, I just wanted to write down an argument that sounded clearer to me. That's all. $\endgroup$ – stressed out Jul 12 '18 at 6:41
  • $\begingroup$ Sorry, I didn't make my definitions clear. I am using the definition of $A$ being open as being an element of the collection of open sets generated by open balls from the metric of $X$. The definition of $\partial A$ that I am using is $\partial A = \bar{A} \setminus \text{Int}(A)$, where $\displaystyle \bar{A} = \cap_{\text{closed sets $C$ containing $A$}} C$ and $\displaystyle \text{Int}(A) = \cup_{\text{open sets $U$ contained in $A$}} U$. $\endgroup$ – Frederic Chopin Jul 12 '18 at 18:19

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