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$(X, \mathcal{T})$ is a topological spaces. $X$ is Hausdorff. $A,B$ are compact, disjoint sets in $X$.

Suppose $X$ is Hausdorff and $A \subset X$, $B \subset X$ are compact and disjoint. Consider an arbitrary point $a_i \in A$. Since $A \cap B = \emptyset$, $a_i \in X - B$. Let $b_i \in B$. Clearly, $a_i \neq b_i$. Since, $X$ is Hausdorff, there exists an open neighbourhood $U(a_i)$ of $a_i$ and an open neighbourhood $V(b_i)$ of $b_i$ such that $U(a_i) \cap V(b_i) = \emptyset$. \ The set $\mathcal{V} = \{V(b_i) \mid b_i \in B\}$ is an open cover of $B$. By compactness of $B$, there exists a finite subcover of $B$ contained in $\mathcal{V}$, say $\{V(b_1),V(b_2), \dots ,V(b_n)\}$. It follows that $B \subset \bigcup_{i=1}^n V(b_i)= V$. Clearly $V \in \mathcal{T}$. Similarly, there exists $U \in \mathcal{T}$ such that $A \subset \bigcup_{i=1}^n U(a_i) = U$.

Now, how do I show $U \cap V = \emptyset$ ?

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There is no reason that $U$ and $V$ in your construction should be disjoint since you don't know whether $U(a_i) \cap V(b_j) = \emptyset$ for $i \neq j$. Try this: Show first that for every $a \in A$ there are open sets $U(a)$ and $V(a)$ such that $a \in U(a)$, $B \subseteq V(a)$ and $U(a) \cap V(a) = \emptyset$ using a similar argument as in your earlier question. Extract a finite subcover $A \subseteq U(a_1) \cup \cdots \cup U(a_n)$ and set $U = U(a_1) \cup \cdots \cup U(a_n)$ and $V = V(a_1) \cap \cdots \cap V(a_n) \supseteq B$. –  Martin Feb 23 '13 at 9:57
    
Please do not delete your questions after receiving an answer. –  Martin Feb 24 '13 at 6:01

2 Answers 2

I think what the questioner is after is a Corollary of the following result which is 3.5.6 on p.84 of the book Topology and Groupoids.

3.5.6 Let $B,C$ be compact subsets of $X,Y$ respectively and let$\mathcal W$ be a cover of $B \times C$ by sets open in $X \times Y$. Then $B,C$ have open neighbourhoods $U,V$ respectively such that $U \times V$ is covered by a finite number of sets of $ \mathcal W$.

The proof starts with the case $B$ is a singleton, and then proceeds to the general case.

Corollaries of this are:

  1. The product of compact spaces is compact.

  2. Let $B,C$ be compact subsets of $X,Y$ respectively and let $W$ be an open subset of $X \times Y$ containing $B \times C$. Then $B,C$ have open neighbourhoods $U,V$ respectively such that $U \times V \subseteq W$.

  3. If $B,C$ are disjoint compact subspaces of the Hausdorff space $X$, then $B,C$ have disjoint open neighbourhoods.

  4. A compact subset of a Hausdorff space is closed.

Edit March 1: Actually there are some lessons for a beginner to derive from the answers and comments.

It is almost always useful to look for the simplest possible case to get started. Here one first tries for both $A,B$ singletons; but that is no good because the result is just the usual Hausdorff condition. So one next tries $A$ compact and $B$ a singleton, and this, as the solutions show, gets one started.

Another method is to consider other possible outlooks on the problem, even if one already has a solution. Thus a condition equivalent to the usual Hausdorff condition on $X$ is that the diagonal $$\Delta_X= \{(x,x)| x \in X\} \;$$ is closed in $X \times X$. Further $A,B$ disjoint is equivalent to $(A \times B) \cap \Delta_X= \emptyset$. Putting these together, one is led to something like a special case of 2. above.

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The OP is clearly after (3), which has a straightforward direct proof; there’s no need to faff about with products. –  Brian M. Scott Feb 23 '13 at 12:23
    
And what would that straightforward proof be? –  Fatsho Feb 23 '13 at 13:16
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@Brian: Some of us like to see the utility of a general result in terms of its corollaries. It is a matter of personal taste, perhaps. –  Ronnie Brown Feb 24 '13 at 11:05
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I prefer first and foremost to answer the question in a way useful to the querent. I’m quite happy to see the utility of a general result in terms of its corollaries, when the general result is actually the point. –  Brian M. Scott Feb 24 '13 at 16:32
    
@Fatsho: Martin’s comment gives you an outline of the argument. –  Brian M. Scott Feb 24 '13 at 16:33

I have a proof. I describe intuitively here.

Pick $a\in A$, for each $b\in B$ we can find $a\in U_b$, $b\in V_b$ open such that $U_b,V_b$ disjoint. The set of all $V_b$ is an open cover of $B$, we find a finite subcover of $V_{b_k}$, and we take intersections of $U_{b_k}$, it a open neighborhood of $a$ such that it does not intersect the union of $U_{b_k}$. Do so for every point in $A$, we will get again an open cover of $A$, and obtain a subcover from it. Take the intersection of the subcover of $A$ to be the final open neighborhood, and the union of corresponding finite subcover of $B$ to be the final open neighborhood, they do not intersect and caontain $A$ and $B$ correspondingly.

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