Prove that given collection of subset is a topology.

35 Views Asked by At

Problem

Let $X$ be a set ; let $\tau_c$ be the collection of all subsets $U$ of $X$ such that $X-U$ either is countable or is all of $X$. Prove that $\tau_c$ is a topology on $X$.

Attempt

1) $X,\phi \in \tau_c $

2) for any set $ {U_\alpha} \subset \tau_c $ where $I$ is the indexed set, $X- \bigcup \limits_{\alpha \in I} U_\alpha $ = $ \bigcap \limits_{\alpha \in I } X- U_\alpha$. If for all $\alpha \in I ,U_\alpha=\phi$ ,we are done. If there are at least one $U_\alpha \neq \phi$, say $U_o$.Then $X-U_o$ is countable. Then $ \bigcap \limits_{\alpha \in I } X- U_\alpha \subset X-U_o$

3) similarly for intersection.

Doubt

Why $X-U_o$ is considered countable?

What if X is some uncountable set.

Any help or suggestion will be appreciated .