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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?

Counterexample:

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 $X.$

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  • $\begingroup$ They mean the set of all such neighborhoods for $n\in\mathbb Z_+$. $\endgroup$ – Matt Samuel Dec 7 '17 at 5:05
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Take any finite subcollection of such neighborhoods, it will be $\{\{n_k\}\times Y~|~k=1,2,\cdots,m\}$. Then it's union is $\{n_1,n_2,\cdots,n_m\}\times Y\neq \mathbb{Z}^{+}\times Y$.

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