# A set is compact in complement topology iff closed in standard topology

Let $$\mathbb{R^n}$$ be a topological space where we define topology $$\tau$$ as follows: a subset $$A \subset \mathbb{R^n}$$ is closed if it is compact in the standard (Euclidean) topology, or if $$A = \mathbb{R^n}$$ (This means that open subsets in $$\tau$$ are those whose complement is compact in standard topology.)

Show that $$F \subset \mathbb{R^n}$$ is compact in $$\tau$$ $$\Leftrightarrow$$ $$F$$ is closed in standard topology.

My proof:

($$\Rightarrow$$)
Let $$F$$ be compact in $$\tau$$. Let $$U$$ be a open (in $$\tau$$) cover of $$F$$. Because $$F$$ is compact there are open sets $$U_1, U_2, \ldots , U_n$$ so that $$F = U_1 \cup U_2 \cup \ldots \cup U_n$$.
For each $$U_i$$ is $$U_i^c$$ compact in standard topology. A union of a finite number of compact sets is compact. So $$F$$ is compact in standard topology. Because $$\mathbb{R}$$ is a metric space, a compact set $$F$$ is closed. (In standard topology.)

($$\Leftarrow$$)
Let $$F$$ be closed in standard topology. Here I am a bit lost as for how to continue. We know that $$F^c$$ is open (in s.t.) Let $$U$$ be an open cover for $$F$$ in $$\tau$$. Can we find a finite number of open sets that cover $$F$$? Those need to have compact complements in standard topology. Can we find a finite number of compact sets that make up a whole $$F$$?

There was a similar question asked a while ago, but I did not completely understand the answer.

• The set $\mathbb{R}^n$ is open in $\mathbb{R}^n$, so the distinction in the definition is unnecessary. Also, are you sure this even defines a topology? The infinite intersection of open sets is not open, so in this definition, an infinite intersection of closed sets need not be closed; but this must hold in any topological space (pretty much by definition). – SvanN Dec 20 '18 at 17:52
• It is true that $\mathbb{R^n}$ is open, but as I am defining topology using closed sets I need to make sure that the whole space is also in the topology. And I couldn't cover that case without explicitly saying that $A$ is closed if it is equal to $\mathbb{R^n}$. Because $\mathbb{R^n}$ is not compact in Euclidean topology. By definition of topology finite union of closed and infinite intersection of closeds need to be closed. This is true in this case. – Coupeau Dec 21 '18 at 14:39
• I think you're mixing up the terms 'closed' and 'compact'. No, you do not need the distinction: $\mathbb{R}^n$ is open in the usual topology on $\mathbb{R}^n$, hence it is closed in your new "topology". But as I indicated, and as jgon explained below, you do not have a topology. – SvanN Dec 21 '18 at 15:45
• OOHH SORRY! This is totally my mistake. I Wrote the definition of topology wrong. It should say$\mathbb{R^n}$ is closed in $\tau$ if it is COMPACT in standard topology or $A = \mathbb{R^n}$ – Coupeau Dec 22 '18 at 11:08

The open subsets of $$\Bbb{R}^n$$ are not closed under arbitrary intersections: $$\bigcap_{n=1}^\infty \left(\frac{-1}{n},\frac{1}{n}\right) = \{0\},$$ so the open subsets of $$\Bbb{R}^n$$ cannot be the closed sets of a topology.
Instead, the question you've linked suggests you should be defining your topology as $$U$$ is open in $$\tau$$ if $$U^C$$ is compact in the usual topology, plus the null set of course.
Then the question you need to answer is why this definition of $$\tau$$ gives a topology.
• Thank you! After reading this and the comments I saw that I made a mistake forming the question. Yes, it should say that a set $A$ is closed if it is COMPACT in the usual topology. Complementing this we get the definition of topology with compact complements, which is indeed, a topology. – Coupeau Dec 22 '18 at 11:15