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I have a question (I guess) regarding my proof below. I have a feeling it's not correct. Would someone kindly verify?

Let $(X,d)$ be a metric space, $a \in X$, $K \subseteq X$ such that $K$ is compact and $a \notin K$. Show that there there exists a neighborhood $V$ of $a$ and an open set $W \supseteq K$ such that $V \cap W = \emptyset$.

Proof. Let $(X,d)$ be a metric space, $a \in X$, $K \subseteq X$ such that $K$ is compact and $a \notin K$.Then, $a \in K^c$. [Note that $K$ compact $\Rightarrow$ $K$ closed $\Rightarrow K^c$ open.] So, $a$ is an interior point of $K^c$. Let $V$ be a neighborhood of $a$, then $N_V(a) \subseteq K^c$. Let $W$ be an open set such that $K \subseteq W$. Now, there are two possible cases:

  1. If $a \in W-K$, then clearly $V \cap W \neq \emptyset$. (Since $W$ contains $a$ and is open, so it also contains $V$.

  2. If $a \in W^c$, then $a \notin W$. So, we can find a neighborhood $V$ of $a$ such that $V \cap W = \emptyset.\ □$

Thank you for the help!

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    You never defined $W$. – 2017-02-28
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    @user399601 I edited it. Thank you for that! – 2017-02-28
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    $W$ can't be arbitrary (in particular, this proof doesn't work). For one thing, it can't contain $a$. The idea will be to use the metric to make sure it can't even get close to $a$. – 2017-02-28
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    PIck neighbourhoods $U_x$ of $x \in K$, $V(x)_a$ of $a$, that are disjoint. Finitely many $U_x$ cover $K$, their union is $W$ ,the finite intersection of the corresponding $V(x)_a$ disjoint from $W$. You can define the sets because we have a metric: $r_x = d(a,x) > 0$. $U_x = B(x, \frac{r_x}{2}), U(a)_x = B(a, \frac{r_x}{2})$ will do. – 2017-02-28
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    You can also forget about the compactness, just use $K$ is closed. Then apply "metric implies regular". – 2017-03-03
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    @HennoBrandsma So my professor wanted us to use compactness. I figured it out, thank you for your help! And I should have been more clear about my background, we are using Baby Rudin so we haven't really seen regular spaces. He did mention them for a few seconds. – 2017-03-03

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I think the idea of this proof is OK, but there are a few assumptions you should make explicit. First, you should explain how you can make neighborhoods of $a$ that are small enough to miss $K$. I'm not quite sure I see exactly how the cases work, but I think we can make it work. We need to make some smart choices about how to define the neighborhoods of $a$ and $K$.

We define a useful function $d$ by $d(x) = $ the infimal distance from $x$ to $K$. Now since $K$ is compact, $K$ contains all limit points, as $K$ is closed, so $d(a) > 0$. Now take a disk of radius $d(a)/3$ around $a$ as the neighborhood of $a$. Next, cover $K$ by putting a disk of the same radius over every point. The union of these open sets can be your neighborhood of $K$. The neighborhoods will not intersect by triangle inequality.

Our proof doesn't need compactness, since arbitrary unions of open sets are open, so you can actually scratch this part all together!

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    But why would you take the disk of radius $d(a)/3$ as the neighborhood of $a$? When you say, "cover $K$ by putting a disk of the same radius...", we do not have to specify what this radius is, do we? Lastly, how did the triangle inequality come into the picture? – 2017-02-28
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    @CuriousMathStudent I'll answer your questions kind of out of order, but I think it will help. First, I wanted a number so that when I take a disk around $a$ and the points of $K$, I would be sure they would not intersect. I guess I could have taken $d(a)/2$ since sets are open, but taking $d(a)/3$ makes a clearer mental picture, for me at least. The triangle inequality works in any metric space, and it is the way in which we prove that these things actually do not intersect, and you should try to flesh this out. Lastly, by 'the same radius' I mean the same number $d(a)/3$. – 2017-02-28
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    I apologize if I'm asking too many questions. I will think about it more and see if make more sense in my head in the morning. – 2017-02-28
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    @CuriousMathStudent suppose an element x belongs to V and W-as Alfred Yerger defined (that is in a neighbourhood of some y in W), there will be a contradiction due to triangle inequality – 2017-02-28
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    @AlfredYerger By triangle inequality, you mean the following: $d(x,z) \leq d(x,y) + d(y,z)$. Right? Because I still don't see how it fits in.. – 2017-02-28
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    @CuriousMathStudent That's right. The idea is that we have chosen balls small enough that they cannot possible intersect. – 2017-03-02
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We need to find a $W$ such that $W\cap V=\emptyset$ in order to complete the proof. To this end we should use the fact that we have a metric, not just a hausdorff topology.

Hence since $V$ is open, there exists some $\varepsilon >0$ such that $B_\varepsilon (a)\subseteq V$, and then $B_{\frac{\varepsilon}{2}} (a)\subseteq\overline{B}_{\frac{\varepsilon}{2}}(a)\subseteq X\setminus K$. Redefine $V=B_{\frac{\varepsilon}{2}} (a)$, and let $W=X\setminus \overline{B}_{\frac{\varepsilon}{2}}(a)$.

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    Hausdorff is enough., metrics is overkill. – 2017-02-28
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    @HennoBrandsma Could you please elaborate on "Hausdorff in enough"? I approached the proof in a different way and I have a feeling that I need to use the definition of a Hausdorff space, but the definition doesn't seem to fit. Thank you! – 2017-03-02
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    @CuriousMathStudent look at my proof sketch in the comments on the question. There I only use Hausdorff – 2017-03-02
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    @HennoBrandsma I disagree that metrics are overkill when the problem says to work in a metric space, and the OP may not know that metric spaces satisfy the separation axioms. – 2017-03-02
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    Your post answer seems to suggest you have to use the metric not just Hausdorff. If the OP knew about Hausdorff It *would* have been enough too. – 2017-03-03
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    So my professor didn't really define any metric, he just took (perhaps, constructed is a better word) neighborhoods of two distinct point and showed they are distinct. He "sneaked" in the Hausdorff property in the proof. – 2017-03-03