Does a metric space need to be a Hausdorff space?

Question

I need to prove that:

Let $X$, $Y$ be metric compact spaces, $f:X \to Y$ be continuous and bijective. Then $f$ is an homeomorphism.

But I have been investigating, and there's no such theorem, because it says that $Y$ needs to be a Hausdorff space.

Can someone explain me why $Y$ needs to be a Hausdorff space?

Thanks.

It is known that metric spaces are always Hausdorff, but that requires proof. Hausdorffness is not part of the definition of a metric space. — Geoffrey Trang, Mar 23 '24 at 02:54
If $x,y$ are distinct points in metric space then for $\epsilon:=\frac{1}{2}d(x,y)>0$ we have $B(x,\epsilon)$ and $B(y,\epsilon)$ as open separations of the points $x,y$, so metric space is Hausdorff. — FShrike, Mar 23 '24 at 03:05

André Caldas · Answer 1 · 2024-03-23T19:22:36.603

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There are two questions:

Why is a metric space Hausdorff.
Why does $Y$ have to be Hausdorff for the theorem to work.

I will answer item "2". Item "1" is very easy.

Notice some interesting and obvious stuff about compact subsets:

If you refine the topology adding more open sets, a compact set might cease being be compact.
If you weaken the topology, by removing open sets, the compact sets never cease to be compact.

Now, notice how this works for closed spaces:

If you refine the topology, you are adding more open sets... obviously, adding more closed sets. What was closed never ceases being close.
If you weaken the topology, however, removing some open sets from the topology makes that their complements cease being closed! Some closed sets cease being closed.

Closed sets and compact sets have these antagonistic properties.

For a continuous function, the inverse image of a closed set is still closed. This is not true for a compact set.

On the other hand, the direct image of a compact set is compact. But this is not true for closed sets.

However, in a compact Hausdorff space, a subset being compact is exactly the same as being closed. So, you get the "best" of both worlds!

A continuous bijection is a homeomorphism when it takes closed sets to closed sets. Well... it does take compact sets to compact sets. If compact and closed are just the same in this context, then a continuous function is always closed.

edited Mar 23 '24 at 19:22

answered Mar 23 '24 at 04:34

André Caldas

5,901

It's not obvious to me that compact sets remain compact when the topology weakens. What if you happened to get rid of just the open sets that were needed for a finite subcover if some open cover? – MJD Mar 23 '24 at 13:27
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@MJD Then the remaining open sets in the family either still cover the space (in which case they have a finite subcover) or they no longer do (in which case you don't have an open cover at all anymore). See https://math.stackexchange.com/questions/3748942/clarifying-why-compactness-in-a-topology-implies-compactness-in-a-coarser-topol for more details. – Carmeister Mar 23 '24 at 15:00
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$X$ doesn't need to be Hausdorff, $Y$ does. – Chad K Mar 23 '24 at 15:47
@ChadK: And $Y$ does not "need" to be compact. But $X$ is in fact Hausdorff and $Y$ is in fact compact, because they are homeomorphic. :-) – André Caldas Mar 23 '24 at 19:03
@MJD: For every finite open cover... well, if you take an open cover in the weaker topology, it is still an open cover in the original topology. Now, compactness implies the existence of a finite subcover. – André Caldas Mar 23 '24 at 19:05
@MJD: If you get rid of the open sets used in the finite subcover, the original cover will not be open anyways. – André Caldas Mar 23 '24 at 19:06
@AndréCaldas: There's a difference between concluding something as a corollary of a theorem and assuming it as a condition for the theorem. $X$ doesn't need to be assumed Hausdorff and $Y$ doesn't need to be assumed compact to apply the theorem. – Chad K Mar 23 '24 at 19:11
@ChadK: oops... I didn't spot my mistake. I will fix it, thank you! – André Caldas Mar 23 '24 at 19:21

score 3 · Answer 2 · answered Mar 23 '24 at 06:02

We have more generally the following fact.

Let $X$ be compact and $Y$ Hausdorff. If $f: X \to Y$ is continuous and bijective, then $f$ is a homeomorphism.

We can prove this by showing that $f$ is a closed map, making it a homeomorphism. Let $C$ be an arbitrary closed set in $X$. Since $X$ is compact, then $C$ is compact [1]. Since $f$ is continuous, the image $f (C)$ is compact in $Y$ [2]. Since $f (C)$ is a compact subspace of a Hausdorff space, then $f (C)$ is closed [3], as desired.

It is easy to see how your statement follows. Since $X, Y$ are both compact metric spaces, then $X$ is compact, and $Y$, being a metric space, is Hausdorff [4]. The fact above applies to show that any continuous and bijective $f: X \to Y$ is a homeomorphism.

Proof of some basic facts:

[1] Closed subspace of compact space is compact.

Let $X$ be compact and $A \subseteq X$ be closed. Suppose $\{ U_\alpha \}$ is a collection of open sets of $X$ that covers $A$. Consider the collection $\{ U_\alpha \} \cup \{ X - A \}$, which is an open cover of $X$. Since $X$ is compact, there is a finite subcover $\{ U_1, \dots, U_N \}$. The collection $\{ U_1, \dots, U_N \} - \{ X - A \} \subseteq \{ U_\alpha \}$ is finite and covers $A$.

[2] Continuous image of compact set is compact.

Suppose $A$ is compact, and let $f$ be continuous. Suppose that $\{ V_\alpha \}$ is an open cover of $f (A)$. Since $f$ is continuous, the collection $\{ f^{-1} (V_\alpha) \}$ is an open cover of $A$. Since $A$ is compact, there is a finite collection $\{ f^{-1} (V_1), \dots, f^{-1} (V_N) \} \subseteq \{ f^{-1} (V_\alpha) \}$ that covers $A$. The collection $\{ V_1, \dots, V_N \} \subseteq \{ V_\alpha \}$ is therefore a finite open cover of $f (A)$.

[3] Compact subspace of Hausdorff space is closed.

Suppose $X$ is Hausdorff, and let $A \subseteq X$ be compact. We show that $X - A$ is open. Let $x \notin A$. For each point $y \in A$, we have that $x \ne y$, so that there exist disjoint neighborhoods $U_y$ of $x$ and and $V_y$ of $y$, since $X$ is Hausdorff. The collection $\{ V_y \}$ is an open cover of $A$; since $A$ is compact, there is a finite collection $\{ V_{y_1}, \dots, V_{y_N} \} \subseteq \{ V_y \}$ that covers $A$. The neighborhood $U = \bigcap_{n = 1}^N U_{y_n}$ of $x$ is disjoint from $A$.

[4] Metric space is Hausdorff.

Let $X$ be a metric space, and $x \ne y \in X$. Therefore, $d (x, y) = \delta > 0$. Consider the neighborhoods $U = B_d \left( x, \frac{1}{2} \delta \right)$ of $x$ and $V = B_d \left( y, \frac{1}{2} \delta \right)$ of $y$. They are disjoint: if $z \in U \cap V$, then $d (z, y), d (z, x) < \frac{1}{2} \delta \implies d (x, y) \le d (x, z) + d (z, y) < \delta$, which is a contradiction.

@BobDobbs I wrote them, but they are of course basic facts that can be found in any introductory text on topology. — K. Jiang, Mar 23 '24 at 17:20

Sat · Answer 3 · 2024-03-23T04:03:13.230

So in metric spaces, basic open sets are open balls, right?

Now pick two points $a\neq b$ from $Y$, there you can define distance between them in metric space, say it is $\delta (>0)$, and now construct the open balls $B(a;\frac{\delta}{4})$ (centered at $a$) and $B(b;\frac{\delta}{4})$, now is there anything inside intersection? If there is something, then you'll have an absurd inequality $\delta\leq \frac{\delta}{2}$!!!

score 0 · Accepted Answer · answered Mar 23 '24 at 13:45

I understood the question to be: why must $Y$ be Hausdorff for this theorem to hold? Put another way, why does the theorem fail if we merely assume $Y$ is compact and not both compact and Hausdorff?

Given any compact metrizable space $X$ with at least two points, let $Y$ have the same underlying set with the indiscrete topology $\{\emptyset,Y\}$. Then the identity map is bijective and continuous, but this map is not a homeomorphism.

score 0 · Answer 5 · answered Mar 23 '24 at 19:34

Another try... so maybe we can pull ourselves more and more out of the box. :-)

I will use the assumptions that $X$ is compact and $Y$ is Hausdorff.

Notice that a bijection means that every point of $X$ is identified with every point of $Y$. But you can think further... every subset of $X$ corresponds to exactly one subset of $Y$. And a family of subsets of $X$ is in bijection with the corresponding family of subsets of $Y$.

In practice, $X$ and $Y$ are the same, and the function $f$ is the identity. We just regard $x$ and $f(x)$ as being the same thing.

So, when you say that \begin{align*} \textrm{id}: (X, \tau) &\rightarrow (X, \gamma)\\ x &\mapsto x \end{align*} is continuous, what you are saying is that $$\gamma \subset \tau.$$ Notice that $\gamma$ is Hausdorff and $\tau$ is compact. When you add open sets to a Hausdorff topology, it does not cease being Hausdorff! And when you remove open sets from a compact topology it does not cease being compact.

So, actually, we have that both are Hausdorff and compact.

I will prove that:

If $\tau$ is compact Hausdorff, you cannot weaken it without loosing the Housdorff property, and you cannot make it stronger without loosing compactness.

Notice that in a compact Hausdorff space, being closed and being compact is the same thing. This is the fact we need.

So, if you weaken the topology, it means some closed set ceased being closed. But it did not ceased being compact. This means the topology is no longer Hausdorff.

On the other hand, if you add to the topology to make it stronger, you have a new, closed set that is not compact. Because if a set is not compact, it cannot be compact in a stronger topology. This means you have a closed set that is not compact. Therefore, the stronger topology may not be compact. (closed subsets of compacts are compact)

This is the compact-Hausdorff rigidity. You cannot make the topology stronger or weaker without loosing compactness, in the former case, or the Hausdorff property in the later.

Any stronger topology is still Hausdorff, but no longer compact.

Any weaker topology is still compact, but no longer Hausdorff.

So, $\gamma = \tau$ and $\textrm{id}$ is a homeomorphism.

Does a metric space need to be a Hausdorff space?

5 Answers5