[Math] Let $(Z,T)$ be the set of integers with the cofinite topology. Find the limit points of the sets $A=\{1,2,3\}$ and $E$ (even integers)

general-topology

Also, compute the closure and the interior of each set.

I have tried proofs for finding the limit points, but I just want to check my reasoning is enough and having trouble with the closure and interior. Thanks.

  1. Set $A=\{1,2,3\}$
    $A$ is finite, so it is closed. Therefore $A$ contains any limit points it could have. The open neighbourhood $\{x\}\cup(Z-A)$ contains only $x$ from $A$, thus $A$ has no limit points.

  2. $E$ is the even numbers.
    $E$ is infinite, therefore every open neighbourhood of $E$ must contain infinitely many points of $E$ for the complement to be finite, so we can conclude that every point in $Z$ is a limit point of $E$. So we can say that $E$ is dense.

Is this a solid answer?

Also, what would my next step be in finding the closure and interior?
Thanks

The closure is the set $A\cup A'$ consisting of all its limit points; would $A\cup A'$ be just the integers $Z$ or is that completely wrong?

Best Answer

A closed set in $Z$ is finite (or the whole set $Z$). Thus the only closed set containing $E$ is $Z$ and therefore the closure of $E$ is $Z$: the closure of a set is the intersection of all closed sets containing it.

The interior of a set is the largest open set contained in it. So, let $U$ be an open set, $U\subseteq E$; then $Z-U$ is closed, but it's clearly infinite, because $Z-U\supseteq Z-E$. The only infinite closed set in $Z$ is $Z$; therefore $Z-U=Z$ and $U=\emptyset$.

Similarly, $A$ contains no nonempty open set and therefore the interior of $A$ is again empty. Of course $A$ is closed, by definition of cofinite topology.

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