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This problem is from section 27 in Munkres' book on topology.

Let $\mathcal{T}$ and $\mathcal{T}'$ be two topologies on the set $X$; suppose that $\mathcal{T}'\supset \mathcal{T}$. What does compactness of $X$ under one of these topologies imply about the compactness under the other?

Here is my attempt at answering this question.

Case (1). Suppose $X$ is compact under $\mathcal{T}$. Since $X$ is compact, for every open covering $\mathcal{A}$ of $X$, there exists a finite sub covering of $X$, that is $\cup _{i = 1} ^{n} A_i = X $. Since $\mathcal{T}$ is coarser than $\mathcal{T}'$, the open sets contained in $\mathcal{T}$ are contained in $\mathcal{T}'$. It follows that every open covering of $X$ in $\mathcal{T}$ is an open covering of $X$ in $\mathcal{T}'$ and every finite subcover of these open covers in $\mathcal{T}$ of $X$ are also in $\mathcal{T}'$. Therefore, the compactness of $X$ under the coarser topology tells us that $X$ is also compact under the finer topology.

Case (2). Suppose $X$ is compact under $\mathcal{T}'$. Let $\mathcal{A}$ be an arbitrary open covering of $X$ by sets in $\mathcal{T}$. Since $\mathcal{T}'$ is finer than $\mathcal{T}$, this open covering of $X$, $\mathcal{A}$ must be in $\mathcal{T}'$. Since $X$ is compact under $\mathcal{T}'$, $\mathcal{A}$ must have a finite subcovering. It follows that this finite subcovering for $X$ must be in $\mathcal{T}$ so $X$ is compact under $\mathcal{T}$ as well. Therefore, the compactness of $X$ under the finer topology tells us that $X$ is also compact under the coarser topology.

Initially, I thought case (2) would not imply compactness in the coarser topology but I cannot think of a counter example. Is my work correct? If not, can someone help me see the flaw in my reasoning?

Nidia
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3 Answers3

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In a coarser space, more sets are compact, essentially because there are fewer open covers to need finite subcovers. That is, if a set is compact in the finer topology then it is compact in the coarser topology. Your proof of this fact is correct.

In a finer space, fewer sets are compact, essentially because there are more open covers to need finite subcovers. The extreme example is the discrete and trivial topologies on an infinite set. In the discrete topology, there is an open cover by singletons, which has no finite subcover. On the other hand, in the trivial topology, all open covers are finite.

Ian
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On case(1) you reason that every covering of $X$ by sets belonging to $\mathcal T$ (hence belonging to $\mathcal T'$) have a finite subcovering. However, that is not enough for $X$ equipped with topology $\mathcal T'$ to be compact. For that this must be true for every covering by sets that belong to $\mathcal T'$.

E.g. let $\mathcal T$ denote the indiscrete topology and let $\mathcal T'$ denote the discrete topology. Then $(X,\mathcal T)$ is compact, but if $X$ is not finite then the finer $(X,\mathcal T')$ is not compact.

drhab
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First, your counterexample is good and shows that if $\tau \subset \tau_1$ and $\mathbb{R}$ is compact under $\tau$, then it is not necessarily compact under $\tau_1$.

Now suppose that $\mathbb{R}$ is compact under $\tau_1$. On this front, you're basically done. Let $\{U_i\}_{i \in I}$ be an open cover of $(\mathbb{R}, \tau)$. Well since $\tau \subset \tau_1$, then $\{U_i\}_{i \in I}$ is an open cover of $(\mathbb{R}, \tau_1)$ as well. Since this space is compact by assumption, then our cover admits a finite subcover $\{U_{i_k}\}_{k=1}^n$. You've gotten here on your own; I'm just reiterating the argument.

Now to finish up, notice that the original open cover $\{U_i\}_{i \in I} \subset \tau$. It follows from basic set theory that any subset of this cover, and in particular our finite subcover $\{U_{i_k}\}_{k=1}^n$ is also a subset of $\tau$.

So since every open cover of $(\mathbb{R}, \tau)$ admits a finite subcover, then we can conclude that it is compact.

jasmine
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