I have a question about the following counterexample, why $U_n = \{n\} \times Y$ has not finite collection covering $X$?

Is it because if you take a finite set $S\subset U_n$ then it's impossible to have $X\subset \bigcup S$ since X is infinite?


Let $Y$ consist of two points; give $Y$ the topology consisting of $Y$ and the empty set. Then the space $X = \mathbb{Z}_{+} \times Y$ is limit point compact, for every nonempty subset of $X$ has a limit point. It is not compact, for the covering of $X$ by the open set $U_n = \{n\} \times Y$ has not finite collection covering $X.$

  • $\begingroup$ They mean the set of all such neighborhoods for $n\in\mathbb Z_+$. $\endgroup$ – Matt Samuel Dec 7 '17 at 5:05

Take any finite subcollection of such neighborhoods, it will be $\{\{n_k\}\times Y~|~k=1,2,\cdots,m\}$. Then it's union is $\{n_1,n_2,\cdots,n_m\}\times Y\neq \mathbb{Z}^{+}\times Y$.


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.