Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the set $X = \{1,2,3\}$, and the topologies $\mathcal{T}_1=\{\emptyset,\{2\},\{1,2\},\{2,3\},X\}$, $\mathcal{T}_2=\{\emptyset,\{1\},\{2,3\},X\}$. Determine whether $\mathcal{T}_1$ and $\mathcal{T}_2$ are compact.

By definition, a set $X$ is compact if every open cover of $X$ has a finite sub-collection that covers $X$. By this definition I see that both $\mathcal{T}_1,\mathcal{T}_2$ are open covers for $X$, since they are finite both have a finite sub-collection that covers $X$. Thus $X$ is compact with respect to both $\mathcal{T}_1$ and $\mathcal{T}_2$. Is this correct?

share|improve this question
    
I guess that it should be quite easy for you to show that every finite topological space is compact. –  Martin Sleziak Jun 21 '11 at 11:18
    
You want to show the right thing, but your second sentence seems to miss the point. A topology itself must always cover the space $X$, since $X$ belongs to it by definition. However, you need to show that any collection of open sets covering $X$ has a finite subcollection already covering $X$. Of course, the finiteness of the topologies is the point here. Also: Did you check if $T_1$ and $T_2$ actually are topologies? –  t.b. Jun 21 '11 at 11:20
2  
It is better not to use the notations $T_1,T_2$ in this context, as they often refer to something else in general topology (separation axioms). –  Mark Jun 21 '11 at 11:25
2  
@Mark Since the "T_1" and "T_2" are typed in a different font in the question, I do not think that this will cause confusion, in this context at least. –  Amitesh Datta Jun 21 '11 at 13:04

2 Answers 2

You seem to be saying $X$ is compact because $\mathcal{T}_i$ is a cover with a finite subcover. This is not really the correct argument. You need to show that every open cover has a finite subcover, not just that this holds for one particular open cover.

In this case observe that every open cover is, in particular, a subset of $\mathcal{T}_i$ (simply because an open cover consists of open sets). Since $\mathcal{T}_i$ is finite, this means every open cover of $X$ must be finite. Hence each open cover is its own finite subcover.

More generally, any space with a finite topology is compact. In particular this holds for a finite space: if $X$ is a finite set, then the power set $\mathcal{P}(X)$ is finite, so any topology on $X$ consists of finitely many opens.

share|improve this answer
    
Let me add to the remark made by wildildildlife in the first paragraph: if $(X,{\cal T})$ is any topological space, then the collection of all open sets has a subcover consisting of only one element: $X$! It would clearly be inappropriate to allow every topological space to be compact on the basis of this argument. –  Amitesh Datta Jun 21 '11 at 13:02

Theorem

A finite topological space $X$ is compact.

Proof. If $X$ is a finite topological space and if $\{U_{\alpha}\}_{\alpha\in A}$ is an open cover of $X$ ($A$ is an index set), then we wish to prove that $\{U_{\alpha}\}_{\alpha\in A}$ has a finite subcover of $X$.

Choose, for each $x\in X$, an index $\alpha_x\in A$ such that $x\in U_{\alpha_x}$. This is possible since $\{U_{\alpha}\}_{\alpha\in A}$ is a cover of $X$. We assert that the collection $\{U_{\alpha_x}\}_{x\in X}$ is a finite subcover of $X$. Indeed, it is finite since $X$ is finite and it is a subcover of $X$ by construction.

Therefore, $X$ is compact. Q.E.D.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.