0

I have seen a few other proofs about this. I have written my own, and I just want to verify the error in my own logic.

The question states: Show that if $f:X\rightarrow Y$ is continuous, where $X$ is compact and $Y$ is Hausdorff, then $f$ is closed map (that is, $f$ caries closed sets to closed set).

Let $U$ be a closed set in $X.$ This implies that $U$ is compact since a closed subspace of a compact space is compact (Theorem 26.2 in Munkres). Since the image of a compact space is compact under a continuous map (by Theorem 26.5) we have that $f(U)$ is compact. Since every compact subspace of a Hausdorff space is closed (Theorem 26.3) this implies that $f(U)$ is closed. Therefore $f$ is a closed map.

I believe that the error comes from the idea that we don't know about other closed sets in $Y$. That my proof does not account for ALL closed sets in $Y$. Is this true?

Killaspe
  • 356
  • 4
    Why do you have to know about other closed sets in $Y$? A closed map carries closed sets to closed sets which you demonstrated. – John Douma Oct 04 '21 at 17:41
  • When I looked at other proofs to verify what I did, they started with closed sets in $Y$. So I was wondering if there was a reason that they did that, as opposed to starting with closed sets in $X$. – Killaspe Oct 04 '21 at 17:42
  • Do you have a reference? You leveraged two other theorems in your proof. Is it possible that the other proofs filled in those details and the details involved $Y$? – John Douma Oct 04 '21 at 17:44
  • 5
    Your proof is correct and there is no mistake (except minor typo). – Danny Pak-Keung Chan Oct 04 '21 at 17:46
  • @JohnDouma yes it appears that they were proving some of the theorems that I used. I just skimmed the proof and saw that they were using closed sets in $Y$ and was confused why that would be necessary, and wondered if I needed to do more with the closed sets in $Y$. – Killaspe Oct 04 '21 at 17:53
  • Except for the typo where you swap $A$ for $U$ halfway through the proof, your proof looks correct. – Mark Saving Oct 04 '21 at 18:31
  • @DavidC.Ullrich ya I was copying the theorems from the book where they use $A$, because I realized some people might not have access to the book, and so my proof wouldn't make sense. – Killaspe Oct 04 '21 at 18:44
  • Other random proofs on the internet could be wrong, you know…. Look for one in a published and respected book e.g. – Henno Brandsma Oct 05 '21 at 05:29

2 Answers2

3

There is no error in your proof. You do not need to know all closed subsets of $Y$, it suffices to know that compact subsets are closed.

By the way, your proof is the standard proof. I guess you found this proof as a solution to Munkres' exercise 6 in §26.

Paul Frost
  • 87,968
-1

Let $f: X \to Y$ be continuous with $X$ being compact, and $Y$ Hausdorff. Show for any closed $F \subset X$, $f(F) \subset Y$ is closed. Let $F \subset X$ be closed, as $X$ is compact it is a well known fact that closed subsets of compact need be compact thus $F$ is compact. And $f$ preserves compactness as it is conitnuous, thus $f(F) \subset Y$ is compact and another will known fact is compact subsets of Hausdorff need be closed thus $f(F) \subset Y$ is closed and $f$ is a closed map.

For proofs of our well known facts:

Closed subset of compact is compact: Let $A \subset X$ be closed with $X$ compact and let $\{U_a\}_{a \in A}, A$ some arbitrary indexing set be an open covering for $A$. Since $A$ is closed, then $X \setminus A$ is open and $\bigcup_{a \in A} U_a \cup X \setminus A$ is an open covering for $X$ and by its compactness there exists a finite subset $B \subset A$ such that

$$\bigcup_{a \in B}U_a \cup X \setminus A$$

is a finite sub cover for $X$ implying $\bigcup_{a \in B}U_a$ is a finite sub cover for $A$ making it compact.

Second fact: Suppose $A \subset X$ is compact and $X$ is Hausdorff, we aim to show $X \setminus A$ is open. Let $x \in X \setminus A$ I will find a neighborhood of $x$ disjoint from $A$. By Hausdorffness of $X$, for each $a \in A$, there exists disjoint open set $U_a \ni x, V_a \ni a$. Thus $\{V_a\}_{a \in A}$ is an open covering for $A$ and by its compactness, there exists some finite $B \subset A$ such that $\{V_a\}_{a \in B}$ is a finite sub cover for $A$ so put $V = \bigcup_{a \in B}V_a$ and put $U = \bigcap_{a \in B}U_b$ which is open as finite intersections of open need be open and as for every $a \in A$, $U_a \cap V_a = \emptyset$, one has $U \cap V=\emptyset$ thus we have found a neighborhood of $x$, namely $U$ which is disjoint from $A$ and $X \setminus A$ is open forcing $A$ to be closed.

MyMathYourMath
  • 4,591