Trying to solve a problem in real analysis ($A$ compact and a closed set intersecting) by contradiction using the definition of a limit point.

Question

The question is:

Let $K$ be compact and let $A$ be closed in $\mathrm{\mathbb{R}}^n$, where $K\cap A = \emptyset$. Then, $\exists\epsilon > 0$ so that $\forall x\in K$ it is true that $\,\beta(\epsilon,x)\cap A = \emptyset$

(note that $\beta$ is the ball centered at $x$ with radius $\epsilon$.)

I tried to prove this by contradiction, it went like this.

Proof :
Assume $\,\forall\epsilon>0\;\exists x\in K$ such that $\,\beta(\epsilon,x)\cap A\neq\emptyset$.
Since $x\notin A$ by one of the premises then $\beta(\epsilon,x)\cap A\setminus x \neq \emptyset$ is true. Then by definition of a limit point, $x$ is a limit point of $A$. But since $A$ is closed it contains all of its limit points by definition and thus we have a contradiction.

I went and asked my professor about why this wasn't a valid proof and he told me that from the assumption it doesn't follow that $x$ is a limit point. I still can't wrap my head around why that isn't the case. Could someone please explain? I would be very grateful!

Thank you for taking the time to read my post.

Answer 1

Take $x\in K$. If it was true that, for every $\varepsilon>0$, $\beta(\varepsilon,x)\cap A\ne\emptyset$, then, yes, $x$ is a limit point of $A$. Perhaps that this is what you are thinking. But this is not what you have here. What you do have here is that, for every $\varepsilon>0$, there is some $x_\varepsilon\in K$ such that $\beta(\varepsilon,x_\varepsilon)\cap A\ne\emptyset$. Note that $x_\varepsilon$ depends upon $\varepsilon$. It doesn't have to be the same point for every $\varepsilon$. And this changes everything. All you know is that the open ball $\beta(\varepsilon,x_\varepsilon)$ intersects $A$. Why would that imply that $x_\varepsilon$ is a limit point of $A$?