properties of first countability
One side is true for all topological spaces: if is in converging , then for any open set of , there is some such that , whence . As a result, .
then we obtain a set of nested open sets containing :
Since each is open, its intersection with is non-empty. So we may choose . We want to show that converges to . First notice tat for any fixed , for all . Pick any open set containing . Then . Hence for all . ∎
From this, we can prove the following corollaries (assuming all spaces involved are first countable):
is closed iff every sequence in that converges to implies that .
Conversely, pick any . By the proposition above, there is a sequence in converging to . By assumption . So , which means that is closed. ∎
is open iff every sequence that converges to is eventually in .
First, suppose is open and converges to . If none of is in , then all of is in its complement , which is closed. Then by the proposition, must be in the closure of , which is just , contradicting the assumption that . Hence for some .
Conversely, assume the right hand side statement. Suppose . Then . By the proposition, there is a sequence in converging to . If , then by assumption, is eventually in , which means for some , contradicting the earlier statement that is in . Therefore, , which implies that , or is open. ∎
Suppose first that is continuous, and in converging to . We want to show that converges to . Let be an open set containing . So is open containing , which implies that there is some such that for all , or for all , which means that .
Conversely, suppose preserves converging sequences and a closed set in . We want to show that is closed. Suppose is a sequence in converging to . Then converges to . Since is in and is closed, by the first corollary above. So too. Hence is closed, again by the same corollary. ∎
|Title||properties of first countability|
|Date of creation||2013-03-22 19:09:26|
|Last modified on||2013-03-22 19:09:26|
|Last modified by||CWoo (3771)|