Leave $ (E, tau) $ Being a second space of Hausdorff locally compact. I want to show that there is a $ (K_n) _ {n in mathbb N_0} subseteq E $ such that $ K_n $ It is compact and $ K_ {n-1} subseteq overset { circ} {K_n} $ for all $ n en mathbb N $ Y $ bigcup_ {n in mathbb N_0} K_n = E $.

The question is basically answered in mathoverflow, but there is a part of the test that I do not understand.

As $ E $ he is the second accountant, $ tau $ has an accounting base $ mathcal B $. It's easy to see that $$ mathcal B_c: = left {U in mathcal B: overline U text {is compact} right } $$ is again a basis for $ tau $.

Now, the construction described in the answer is as follows:

Leave $ U_0 en mathcal B_c $ Y $ K_0: = overline {U_0} $. Dice $ (U_0, K_0), ldots, (U_ {n-1}, K_ {n-1}) $, leave $ U_n in mathcal B_c setminus left {U_0, ldots, U_ {n-1} right } $, $ mathcal C $ be a finite sublayer of $ K_ {n-1} $ Y $ K_n: = overline {U_n} cup overline {C} $.

The problematic part is the cover. $ mathcal C $ of $ K_ {n-1} $ (why *sub*cover?). Power $ mathcal C $ Actually be arbitrary. And if we choose? $ mathcal C = E $?