Topology problem about Hausdorff space and compactness

[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.

 

[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.

 

[Jaggis]

But before anyone can help you, you must show as per forum policy what you have tried thus far.

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.

 

[WannabeNewton]

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.

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

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).

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.

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.

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 $X$.

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.

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.

 

[micromass]

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.

 

[Jaggis]

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.

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.

 

[micromass]

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 $C$ .

And what exactly is $L$?

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?

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

 

[Jaggis]

And what exactly is $L$?

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

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

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 $\mathcal{T}_0$?

 

[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.

 

[micromass]

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.

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.

 

[Jaggis]

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.

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.

 

[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.

 

[Jaggis]

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.

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?

 

[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?

 

[Jaggis]

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

You should add $U_0$ $≠$ $∅$ ?

Otherwise, no.

 

[micromass]

You should add $U_0$ $≠$ $∅$ ?

Otherwise, no.

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.

 

[Jaggis]

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.

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.

 

[micromass]

That seems right!

 

[WannabeNewton]

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.

Brilliant! You're done then :)

 

[Jaggis]

Brilliant! You're done then :)

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.

 

[WannabeNewton]

I'm sure you'll do fine mate. I wish you the best of luck, come back sometime and let us know how it went. And be sure to ask more questions if you have them!