# Topology problem about Hausdorff space and compactness

1. May 19, 2013

### Jaggis

Would anyone have ideas on how to solve the following problem?

Let (X, τ) be a Hausdorff space and τ0 = {X\K: so that K is compact in (X, τ)}

Show that:

1) τ0 is a topology of X.

2) τ0 is rougher than τ (i.e. τ0 is a genuine subset of τ).

3) (X, τ0) is compact.

This was a question in a recent exam that I took (I failed). I was especially clueless with 3).

Thanks for any help.

2. May 19, 2013

### WannabeNewton

This is related to what is called the Alexandroff extension or Alexandroff compactification. You'll be happy (or not happy?) to know that the given problems are actually quite simple to solve. But before anyone can help you, you must show as per forum policy what you have tried thus far.

3. May 20, 2013

### Jaggis

To show that τ0 is a topology of X, I tried to show that arbitrary unions and finite intersections of members of τ0 are also members of τ0 and that X and ∅ are included in τ0.

τ includes all members of τ0, since from K is compact and (X, τ) Hausdorff follows that K is closed and therefore X/K is open in (X, τ).
To show that τ0 ≠ τ, I argued that because (X, τ) is Hausdorff, all members of that space must have disjoint neighbourhoods, which would not be possible if all the open sets were defined as X/K, K compact. Therefore, τ should contain more open sets than τ0.
Thus, τ0 is a genuine subset of (or rougher than) τ.

Apparently the compactness of (X, τ0) follows from the fact that since all its open sets are defined as {X/K: K is compact in (X, τ)}, an open neighbourhood for any x∈(X, τ0) will be X/K whose complement K is compact? From this it would follow that (X, τ0) is compact. I don't know how to prove the lemma, however.

4. May 20, 2013

### WannabeNewton

And were you successful? Did you do this part correctly or did you have some concerns regarding this?

This is perfectly fine but let's write it in a more sequential way. Let $U\in \mathcal{T}_{0}$ then $U = X \setminus K$ for some $K\subseteq X$ compact under $\mathcal{T}$. Since $X$ is Hasudorff under $\mathcal{T}$, $K$ is closed under $\mathcal{T}$ so $U = X\setminus K\in \mathcal{T}$. This is exactly what you said but I have just written it in a more sequential manner.

This is actually false. Let $X$ be a finite set with the discrete topology; $X$ is Hausdorff as well as compact. Let $U\subseteq X$ be open then $X\setminus U$ is closed in $X$ and is therefore a compact subset of $X$ so $U = X \setminus K$ for a compact subset $K \subseteq X$ hence if $U$ is open in the discrete topology on $X$ it is also open in the topology $\mathcal{T}_{0}$ on $X$.

Recall that a topological space $X$ is compact if every open cover of $X$ has a finite subcover. So let $\mathcal{O}$ be an open cover of $X$ under $\mathcal{T}_{0}$ and show that it has a finite subcover.

5. May 20, 2013

### micromass

Staff Emeritus
Like WBN already indicated, I think that when the exercise says that "$\mathcal{T}_0$ is rougher than $\mathcal{T}$", then they mean that $\mathcal{T}_0\subseteq \mathcal{T}$, but that they don't ask for it to be a genuine subset.

6. May 20, 2013

### Jaggis

I think you mean this:

Let $C$ be an open cover of ($X$, $\mathcal{T}_{0}$).

For a compact $L$ there exists a finite subcover {$X$\$K$j: $j$∈$J$, # $J$ < ∞}, which is a subset of $C$ .

Therefore the cover of the whole space ($X$, $\mathcal{T}_{0}$) is [$X$\$L$]$⋃${$X$\$K$j: $j$∈$J$, # $J$ < ∞}, i.e. [$X$\$L$]$⋃$[$X$\$K$1]$⋃$ ... $⋃$[$X$\$K$n], which is finite.

However, to me this raises a question:

If we choose $X$ = $ℝ$ and $L$ = $[0,1]$, which is compact.

Now a finite open cover of $ℝ$ would be [$ℝ$\$[0,1]$]$⋃$[$ℝ$\$K$1]$⋃$ ... $⋃$[$ℝ$\$K$n].

But wouldn't this be a contradiction since $ℝ$ isn't compact?

PS. Apologies about the ugly notation. I'm just starting to learn how to add and use the symbols.

7. May 20, 2013

### micromass

Staff Emeritus
And what exactly is $L$?

The space $\mathbb{R}$ under the usual Euclidean topology is not compact. The space $\mathbb{R}$ with $\mathcal{T}_0$ is compact.

8. May 20, 2013

### Jaggis

Any compact set of ($X$,$\mathcal{T}$). $X$\$L$ will be open in ($X$, $\mathcal{T}_0$).

But even in Euclidean topology $\mathcal{T_E}$ one could claim that $\mathbb{R}$ is compact due the fact that its finite open cover is

{$\mathbb{R}$\$[0,1]$, $U$1, ..., $U$n}

as

i) $[0,1]$ is compact in ($\mathbb{R}$,$\mathcal{T_E}$) and it will have a finite open cover $U$1, ..., $U$n.

and

ii) $[0,1]$ is closed in ($\mathbb{R}$,$\mathcal{T_E}$) and thus its complement $\mathbb{R}$\$[0,1]$ is open.

So what makes the difference with $\mathcal{T}_0$?

9. May 20, 2013

### WannabeNewton

Your method is not going to help you prove that the new space is compact. You are approaching it in a wrong way, although your overall idea is in the right ballpark. Let me show you a method that might seem to work and see if you can fix it to get the correct proof. Let $\mathcal{O}$ be an open cover of $(X,\mathcal{T}_{0})$ and let $U_{0}\in \mathcal{O}$. We know that $X\setminus U_{0}$ is compact in $(X,\mathcal{T})$ by definition. Therefore, since $\mathcal{O}$ is an open cover of $X\setminus U_{0}$ by open subsets of $(X,\mathcal{T}_{0})$, there exists a finite subcover $\{U_1,...,U_n\}$ of $X\setminus U_{0}$ hence $\{U_0,U_1,...,U_n\}$ covers all of $X$ therefore $X$ is compact. There is a subtle mistake here, that I have purposefully made for you to point out. Can you tell me what it is? If you can find the mistake you can easily fix it and write down the correct proof.

10. May 20, 2013

### micromass

Staff Emeritus
This is due to a subtle misunderstanding of compactness. What you proved now is that $\mathbb{R}$ has a finite open cover. This is true. In fact, the cover $\{\mathbb{R}\}$ itself is already a finite open cover. But this does not imply compactness. Compactness says that every open cover has a finite subcover. So simply exhibiting an open cover is not enough, you must actually find a finite subcover for every single open cover of the space.

Now, $\mathbb{R}$ is not compact because $\{(n-1,n+1)~\vert~n\in \mathbb{Z}\}$ has no finite subcover.

11. May 20, 2013

### Jaggis

I'm sorry, but I simply don't know what the catch is here. Like I showed before with $ℝ$ and $[0,1]$ in Eucledian (or any) topology, this proof doesn't make sense to me and I don't how to correct it to make it work.

I'm just as clueless with the whole thing as I was in the beginning.

12. May 20, 2013

### WannabeNewton

Can you tell me the definition of compactness that you were taught in class? I don't think I can make the above any easier for you without giving it away I'm afraid so maybe we need to reinforce the concepts in your head.

13. May 20, 2013

### Jaggis

Okay, I may be getting it now.

$X$ will be compact only if there exists a finite subcover for any open cover of $X$.

Any open cover of the new space will include a member $U$ whose complement $X$/$U$ is always compact and will have a finite subcover {$U$1, ..., $U$n}. Thus the whole space will always have a finite subcover $U$ ⋃ {$U$1, ..., $U$n} regardless of the choice of the open cover.

With $ℝ$ and Euclidean topology one can find an open cover of $ℝ$ that has a member whose complement isn't compact and therefore one cannot find a finite open subcover for any open cover of $ℝ$. Therefore $ℝ$ isn't compact with Euclidean topology.

Or maybe I'm just dancing around the problem?

14. May 20, 2013

### WannabeNewton

Yes that is correct with regards to what compactness is. Now can you spot the mistake in the proof I gave in post #9?

15. May 20, 2013

### Jaggis

You should add $U_0$ $≠$ $∅$ ?

Otherwise, no.

16. May 20, 2013

### micromass

Staff Emeritus
The concept "open set" is dependent on the topology. So a set might be open in one topology, but maybe not in the other. The same with the notions of compactness.

So, in your proof you should be careful about which sets are open/compact and in which topology.

17. May 21, 2013

### Jaggis

Perhaps the mistake here is that you jump into thinking that $X\setminus U_{0}$ is compact in $(X,\mathcal{T}_{0})$ just because it's compact in $(X,\mathcal{T})$ without proof.

First, the open sets of $\mathcal{T}_{0}$ will cover $(X,\mathcal{T}_{0})$. Since $\mathcal{T}$ contains $\mathcal{T}_{0}$, an open cover $\mathcal{O}$ of $(X,\mathcal{T}_{0})$ is always an open cover of $(X,\mathcal{T})$ and therefore it will cover $X\setminus U_{0}$ too. Since $X\setminus U_{0}$ is compact in $(X,\mathcal{T})$, it has a finite subcover $\{U_1,...,U_n\}$ which is a subset of $\mathcal{O}$. As the sets of $\{U_1,...,U_n\}$ are open in both spaces, $X\setminus U_{0}$ always has a finite subcover in both spaces and therefore is compact in both spaces.

18. May 21, 2013

### micromass

Staff Emeritus
That seems right!

19. May 21, 2013

### WannabeNewton

Brilliant! You're done then :)

20. May 21, 2013

### Jaggis

OK! Thank you very much for your help, WannabeNewton and micromass.

Topology is still quite tricky business to me, but I hope this exercise prepared me for challenges to come so I can finally pass my course some day.