0
$\begingroup$

I'm studying for a test on general topology and in some old exam the following question has been asked:

For any finite subset $S\subseteq\mathbb{Z}\setminus\{0\}$ we define $U_S:=\mathbb{Z}\setminus S$. Let $\mathcal{T}$ be the set defined by $\mathcal{T}=\{\emptyset\}\cup\{U_S\mid S\subseteq\mathbb{Z}\setminus\{0\} \text{ finite}\}$.

(a) Show that $\mathcal{T}$ defines a topology on $\mathbb{Z}$
(b) Show that $(\mathbb{Z},\mathcal{T})$ is compact

I've managed to do (a). However (b) is not really working out. Can someone give me a good tip on how to solve this, thanks in advance.

$\endgroup$
2
$\begingroup$

Let $\{U_i, i \in I\}$ be an open cover of $X$. Then some open set $U_{i_0}$ must contain $0$, this $U_{i_0}$ must be of the form $U_S$ for some finite $S \subseteq \mathbb{Z}\setminus\{0\}$ by the definition of the topology.

For each $s \in S$ there must be some $U_{i_s}$ that contains $s$ (as we have a cover), and then $\{U_{i_0}, U_s: s \in S\}$ is a finite subcover (any $x$ in $\mathbb{Z}$ is in $S$ and so covered or else it’s in $U_S=U_{i_0}$) of our cover and we are done.

Note that this topology is a weird modification of the co-finite (or finite-closed) topology on $\mathbb{Z}$, modified so that all non-empty open sets contain $0$, so $\{0\}$ is a dense subset of $X$. $X$ is thus only $T_0$, not $T_1$ as the co-finite topology is.

$\endgroup$
1
$\begingroup$

Let $\{A_\lambda\,|\,\lambda\in\Lambda\}$ be an open cover of $\mathbb Z$. If $A_\lambda=\mathbb Z$ for every $\lambda\in\Lambda$, then $\{A_{\lambda_0}\}$ (where $\lambda_0$ is some element of $\Lambda$) is an open subcover. Otherwise, take $\lambda_0\in\Lambda$ such that $A_{\lambda_0}\neq\mathbb Z$. Then $\mathbb{Z}\setminus A_{\lambda_0}$ is finite. So…

$\endgroup$

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.