Let $\tau ,\tau_1$ be two topologies on the set $\mathbb R$ .Suppose $\tau \subset \tau_1$ .What does compactness of $\mathbb R$ under one of these topologies imply about compactness under the other?

My try: Taking ($\mathbb R,$discrete) and ( $\mathbb R$,indiscrete ) is a counterexample .

If $\mathbb R,\tau_1$ is compact then to show that $\mathbb R,\tau $ is compact .let $\cup U_\alpha $ be an open cover of ($\mathbb R,\tau $ ) then it is an open cover in ($\mathbb R,\tau_1$) which is compact thus $\exists $ $U_i:1\geq i\geq n$ in ( $\mathbb R,\tau_1 )$ which covers $\mathbb R$

How to ensure that these are open in $\mathbb R,\tau $ .please help if this proof can be completed or there exists a counter example


1 Answer 1


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.

  • $\begingroup$ i got your point $\endgroup$
    – Learnmore
    May 19, 2015 at 11:01

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