1. ## Metric space question...

I am having difficulty approaching this question... any help would be greatly appreciated! Thank you!

Show that in a metric space the intersection of a compact set and a closed set is compact.

2. Originally Posted by Majialin
I am having difficulty approaching this question... any help would be greatly appreciated! Thank you!

Show that in a metric space the intersection of a compact set and a closed set is compact.

In a metric space a set is compact iff it is closed and bounded.

The finite intersection of closed sets is closed.

Note that if A is your compact set and B is your closed set then

$A \cap B \subset A$

3. Originally Posted by TheEmptySet
In a metric space a set is compact iff it is closed and bounded.
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This is only true in finite dimensional normed spaces. For metric spaces compactness is equivalent to [completeness and totally-boundedness] and sequential compactness which I believe would be more useful in this case

4. In addition, the set of rational numbers, with the metric d(x,y)= |x- y| has closed and bounded sets that are not compact.

5. Originally Posted by HallsofIvy
In addition, the set of rational numbers, with the metric d(x,y)= |x- y| has closed and bounded sets that are not compact.

Perhaps a slightly simpler and sharper example: the reals with the discrete topology (from the discrete metric), and $X=\{1,2,3,4,...\}=\mathbb{N}$ with the inherited topology . X is closed and totally bounded, but not compact.

Tonio

6. The answer to the original question of Majialin doesn't need so subtle concepts as totally bounded. In a Hausdorff space is clear that a closed subset of a compact is again compact.

I am not sure in the non Hausdorff case, I am not used to think in this spaces, it is not the point but, is it also true?

7. It is not serious, but I can answer my question given the argument for the question. It follows from the very definition.

If K is compact and F is a closed subset of K, then any open covering of F can be "extended" to an open covering of K by adding $K\setminus F$. Hence it follows compactness of F, even not assuming ithe space to be Hausdorff.

8. Originally Posted by Enrique2
The answer to the original question of Majialin doesn't need so subtle concepts as totally bounded. In a Hausdorff space is clear that a closed subset of a compact is again compact.

I am not sure in the non Hausdorff case, I am not used to think in this spaces, it is not the point but, is it also true?
Well, every metric space is a paracompact Hausdorff space. So are you considering other spaces?

9. Originally Posted by Enrique2
The answer to the original question of Majialin doesn't need so subtle concepts as totally bounded. In a Hausdorff space is clear that a closed subset of a compact is again compact.

I am not sure in the non Hausdorff case, I am not used to think in this spaces, it is not the point but, is it also true?
Yes, that's a very good suggestion. Any compact set, in a Hausdorff space, is closed and the intersection of two closed sets is closed so the intersection of a compact set and a closed set is compact. Then use the fact that a closed subset of a compact set is closed.

And, of course, any metric space is Hausdorff.

10. Originally Posted by HallsofIvy
Yes, that's a very good suggestion. Any compact set, in a Hausdorff space, is closed and the intersection of two closed sets is closed so the intersection of a compact set and a closed set is compact. Then use the fact that a closed subset of a compact set is closed.

And, of course, any metric space is Hausdorff.
Yes, of course. I only introduced the non Hausdorff case because of my curiosity. And I believe the argument works, more detailedly

if $f
F\subset K$
, $F$ closed and $F=\cup_{i\in I}(V_i\cap F)$,
$V_i$ open in $K$, then $K=\cup_{i\in I}V_i\cup K\setminus F$. If $K$ is compact then there exists $J\subset I$ finite such that $K=\cup_{i\in J}V_i\cup K\setminus F$. Hence $F=\cup_{i\in J}V_i\cap F$, i. e. $F$ is also compact. We do not need HAusdorff as I believed in my first post, then the original problem in a metric space is a particular case, in any topological space the propertry is true, always closed intersected compact is compact (if hopefully I have not missed anything!). Sorry because my previous posts were somehow confusing.