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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.2011-06-21
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    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?2011-06-21
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    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).2011-06-21
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    @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.2011-06-21

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