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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?

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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
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
@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

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.

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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


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.

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