# If X is second-countable, then X is Lindelöf space.

general-topologyreal-analysis

Suppose that $$X$$ has countable basis, then every open covering of $$X$$ contains a countable subcollection covering $$X$$.

Proof

{$$U_i;i\in I$$} is open cover for ($$X,\tau$$) and $$B$$ = {$$V_n;n\in N$$} is countable base for topology $$\tau$$.We have that $$X=\cup_i U_i$$.For every point $$x$$ in $$X$$ let's choose any $$U_i$$ s.t. $$x \in U_i$$.Denote that $$U_i$$ as $$U_x$$.There exists $$V_x \in B$$ s.t. $$x\in V_x\subset U_x$$.Collection of $$V_x$$ is countable.For every $$V_x$$ choose $$U_x$$ s.t. $$V_x\subset U_x$$.Because $$X=\cup_xV_x$$ follows that $$X=\cup U_x$$ and number of $$U_x$$-s are countable.

I understood proof only last part I can't understand.Why we need that number of $$U_x$$ to be countable?

We have already open cover for $$X$$ and found {$$V_x$$} which is countable subcollection covering $$X$$.

{$$U_i;i\in I$$} is open cover for ($$X,\tau$$) and $$B$$ = {$$V_n;n\in N$$} is countable base for topology $$\tau$$.We have that $$X=\cup_i U_i$$.For every point $$x$$ in $$X$$ let's choose any $$U_i$$ s.t. $$x \in U_i$$.Denote that $$U_i$$ as $$U_x$$.There exists $$V_x \in B$$ s.t. $$x\in V_x\subset U_x$$.
Now as these $$\{V_{x}\}$$ which correspond to the $$U_{x}$$'s is a subcollection of $$B$$. Hence is countable as a subset of countable set is countable. Now it's just a matter of indexing. Explicitly we were looking at $$U_{0}=\{U_{x}\in \{U_{i}\}_{i\in I} : x\in V_{x}\subset U_{x}\}$$. correspondingly we have $$B_{0}=\{V_{x}: V_{x}\subset U_{x}\}$$. These sets $$B_{0}$$ and $$U_{0}$$ are in bijection and $$B_{0}$$ being a subset of countable set is countable.
Write out all such $$V_{x}$$'s as a sequence say $$\{V_{1},V_{2},.....\}$$. Correspondingly we have $$\{U_{1},U_{2},....\}$$.
Thus $$\{U_{k}\}_{k=1}^{\infty}$$ is a countable subcover of $$\{U_{i}\}_{i\in I}$$