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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.

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    $\begingroup$ 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). $\endgroup$ – SvanN Dec 20 '18 at 17:52
  • $\begingroup$ 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. $\endgroup$ – Coupeau Dec 21 '18 at 14:39
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    $\begingroup$ 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. $\endgroup$ – SvanN Dec 21 '18 at 15:45
  • $\begingroup$ 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 $A = \mathbb{R^n}$ $\endgroup$ – Coupeau Dec 22 '18 at 11:08
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As SvanN says in the comments, your "topology" doesn't form a topology.

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.

After that, you can try to reread the answer to the linked question, and it should hopefully make more sense.

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  • $\begingroup$ 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. $\endgroup$ – Coupeau Dec 22 '18 at 11:15
  • $\begingroup$ The question is now corrected. $\endgroup$ – Coupeau Dec 22 '18 at 11:23

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