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An open cover $\mathcal U$ of a topological space $X$, is called

  1. An $\omega$-cover, if every finite subset of $X$, is contained in a member of $\mathcal U$.

  2. A $\gamma$-cover if it is infinite and every $x \in X$ belongs to all but finitely many elements of $\mathcal U$.

Can anyoone think of a countable $\gamma$-cover which is not an $\omega$-cover for some Hausdorff topological space $X$?

Thank you!

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No such example exists.

Assume that there were a countable $\gamma$-cover that is not an $\omega$-cover. Then there is some finite set $a_1, \ldots, a_n$ not included in any member of the cover.

By the property of a $\gamma$-cover, there are only finitely many (say $N_1$) sets in the cover not containing $a_1$, and $N_2$ set not containing $a_2$, etc. Then there can be atmost $\sum_{i=1}^n N_i$ sets in the cover that miss at least one $a_i$. But, as there are countably many sets in the cover, there then has to be a set containing all $a_i$, thus contradicting the assumption our cover were no $\omega$-cover.

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  • $\begingroup$ ok understood. Thank you! $\endgroup$ – topsi Feb 20 '14 at 17:08

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