0
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.