Let $X=\mathbb R$ with cofinite topology and $A=[0,1]$ with subspace topology – show $A$ is compact

Question

Let $X=\mathbb R$ with the cofinite topology and $A=[0,1]$ with the subspace topology. I've just proven that every closed subspace of a compact space is compact, and now I'm asked to show that $A$ is compact, but not a closed subset of $X$. (I've shown this second part.)

I know that $\mathbb R$ is compact when given the cofinite topology, so to show $A$ is compact I need to show it is a closed subspace of $X$, but I'm not really sure what the distinction is between closed subspace and closed subset. I know that the open sets in the subspace topology consist of $U=V\cap A$ for $V$ open in $X$, so do I just need to show that $[0,1]$ cannot be written in this form?

Suppose it can be written as $[0,1]=V\cap [0,1]$ for some $V$ open in $X$, so $V=\mathbb R\backslash\{x_1,…,x_k\}$. That would mean that $\mathbb R\backslash\{x_1,…,x_k\}\subset [0,1]$ which is clearly false, so $A$ is a closed subspace of $X$. (I only worked this out while asking the question so basically asking if this is correct now)

EDIT: Just realised this doesn't imply $\mathbb R\backslash\{x_1,…,x_k\}\subset [0,1]$ but actually that $[0,1]\subset \mathbb R\backslash\{x_1,…,x_k\}$, which doesn't lead me anywhere.

Answer 1

There are a few things you seem to be misunderstanding:

1. If $X$ is a compact space and $A\subseteq X$ is closed, then $A$ is compact with the subspace topology, but for the converse to be true, $X$ needs to be Hausdorff, which is not true in your case. (In fact, it is not difficult to show that if $X$ is any topological space, then you can find $X'\supsetneq X$ in which $X$ is not closed, which is compact and $T_1$ as long as $X$ is.)
2. As you have seen yourself, in your case, $A$ is not a closed subspace of $X$, and $X$ is clearly not Hausdorff, so the above criterion is not helpful at all. So no, you don't need to show that $A$ is closed in $X$.
3. What you can do is just check directly that $A$ is not an essentially infinite union of open subsets of $A$, that is, there is no infinite family $\mathcal A$ of open subsets of $A$ (in the subspace topology) such that $\bigcup \mathcal A=A$ and the union of every finite subfamily of $\mathcal A$ is a proper subset of $A$.
4. $U$ is an open subset of $A$ in the subspace topology when $U=V\cap A$ for some $V\subseteq X$ open, not when $U=V\subseteq A$ for some $V\subseteq X$ open (unless $A$ itself happens to be open, in which case the two definitions coincide).
5. In this case, it is not hard to see that if $X$ is any space with the cofinite topology and $A\subseteq X$, then the subspace topology is still the cofinite topology. The cofinite topology is always compact, so $A$ is compact.