Consider the interval $[0,1]$ in $\mathbb{R}.$ Let $\mathcal{B}$ be a subspace topology of $[0,1],$ that is, collection of all sets of the form $[0,1]\cap U$ where $U$ is open in $\mathbb{R}.$ Note that $U = (a,b)$ for some $a<b,$ possibly infinite.

Now, if we add $\{0\}$ into $\mathcal{B}$ and denote the new topology as $\mathcal{B}_0,$ then $\{0\}$ is open in $[0,1]$ with respect to $\mathcal{B}_0,$

Question: Is $[0,1]$ compact in the new topology $\mathcal{B}_0$?

Recall that to show that $[0,1]$ is compact, we need to show that every open covering of $[0,1]$ has a finite subcover.

I think $[0,1]$ is not compact in $\mathcal{B}_0$ because of the newly added open set $\{0\}.$ Consider the covering $$\bigg\{ \{0\}\cup (\frac{1}{n},1]: n\in\mathbb{N} \bigg\}.$$ It is an open covering without finite subcover. Therefore, $[0,1]$ is not compact in $\mathcal{B}_0.$

Is my attempt correct?

  • $\begingroup$ Yes, it is correct. $\endgroup$ – user10354138 Oct 11 '18 at 4:51

I assume that adding $0$ into $\mathcal{B}$ means taking the topology generated by $\mathcal{B} \cup \{0\}$. In any case, $\{0\}$ is open for $\mathcal{B}_0$. Now, since

$$ [0,1] = \{0\} \ \cup \ (0,1] $$

and $(0,1]$ is not compact for the original topology $\mathcal{B}$, there exists a covering $(U_i)_{i \in I}$ of $(0,1]$ by open sets in $\mathcal{B}$ (and hence in $\mathcal{B}_0$) with no finite subcover. Therefore, $\{0\} \ \cup (U_i)_{i \in I}$ covers $[0,1]$ and no finite subcover can be extracted. That is to say, your attempt is correct: you have even identified such a cover for $(0,1]$.

  • $\begingroup$ It's the argument I gave in disguise: if $O$ is new open, its complement is newly closed, so could not have been compact; the witnessing cover plus $O$ are a new witnessing cover for non-compactness of $X$. It depends ultimately on compact subsets of Hausdorff spaces are closed either way. $\endgroup$ – Henno Brandsma Oct 11 '18 at 5:13
  • $\begingroup$ Agreed: your way is shorter and I should have added that if $\tau$ makes $X$ compact Hausdorff, no strictly finer topology makes $X$ compact. I just felt like a concrete approach would show the OP why his constructive argument works. $\endgroup$ – Guido A. Oct 11 '18 at 5:17

You're indeed right. You might want to add an argument that there indeed is no finite subcover (it's not very hard, but your teacher might insist).

In general it's true that if $(X, \mathcal{T})$ is compact Hausdorff and $(X,\mathcal{T'})$ is another topology on the same set with $\mathcal{T} \subseteq \mathcal{T}'$, then $(X,\mathcal{T}')$ is not compact unless $\mathcal{T} = \mathcal{T}'$. Here we have a proper superset (as $\{0\}$ is new open and was not open), so the non-compactness follows from this fact too.

Proof of the fact:

Suppose $(X,\mathcal{T}')$ is compact. Consider the identity map $f(x) = x$ from $(X,\mathcal{T}') \to (X,\mathcal{T})$. As $\mathcal{T} \subseteq \mathcal{T}'$, $f$ is continuous. As the domain is assumed to be compact and the codomain is Hausdorff, a standard theorem says that $f$ is a closed map. As $f$ is a bijection, this means that $f$ is a homeomorphism, hence an open map, and $\mathcal{T}' \subseteq \mathcal{T}$ follows and we have equality of the topologies.

This is sometimes stated as "a compact Hausdorff space is maximally compact" (ie. no stricly finer topology can be compact anymore).


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