properties of first countability

Proposition 1.

Let X be a first countable topological spaceMathworldPlanetmath and xX. Then xA¯ iff there is a sequence (xi) in A that convergesPlanetmathPlanetmath to x.


One side is true for all topological spaces: if (xi) is in A converging x, then for any open set U of x, there is some i such that xiU, whence UA. As a result, xA¯.

Conversely, suppose xA¯. Let {Bii=1,2,} be a neighborhood base around x. We may as well assume each Bi open. Next, let


then we obtain a set of nested open sets containing x:


Since each Ni is open, its intersectionDlmfMathworldPlanetmath with A is non-empty. So we may choose xiNiA. We want to show that (xi) converges to x. First notice tat for any fixed j, xiNj for all ij. Pick any open set U containing x. Then NjBjU. Hence xiU for all ij. ∎

From this, we can prove the following corollaries (assuming all spaces involved are first countable):

Corollary 1.

C is closed iff every sequence (xi) in C that converges to x implies that xC.


First, assume (xi) is in a closed setPlanetmathPlanetmath C converging to x. Then xC¯ by the propositionPlanetmathPlanetmath above. As C is closed, we have xC¯=C.

Conversely, pick any xC¯. By the proposition above, there is a sequence (xi) in C converging to x. By assumptionPlanetmathPlanetmath xC. So C¯C, which means that C is closed. ∎

Corollary 2.

U is open iff every sequence (xi) that converges to xU is eventually in U.


First, suppose U is open and (xi) converges to xU. If none of xi is in U, then all of xi is in its complement X-U, which is closed. Then by the proposition, x must be in the closureMathworldPlanetmathPlanetmath of X-U, which is just X-U, contradicting the assumption that xU. Hence xiU for some i.

Conversely, assume the right hand side statement. Suppose xU=X-X-U¯. Then xX-U¯. By the proposition, there is a sequence (xi) in X-U converging to x. If xU, then by assumption, (xi) is eventually in U, which means xiU for some i, contradicting the earlier statement that (xi) is in X-U. Therefore, xU, which implies that UU, or U is open. ∎

Corollary 3.

A functionMathworldPlanetmath f:XY is continuousPlanetmathPlanetmath iff it preserves converging sequences.


Suppose first that f is continuous, and (xi) in X converging to x. We want to show that (f(xi)) converges to f(x). Let V be an open set containing f(x). So f-1(V) is open containing x, which implies that there is some j such that xif-1(V) for all ij, or f(xi)V for all ij, which means that (f(xi))f(x).

Conversely, suppose f preserves converging sequences and C a closed set in Y. We want to show that D:=f-1(C) is closed. Suppose (xi) is a sequence in D converging to x. Then (f(xi)) converges to f(x). Since (f(xi)) is in C and C is closed, f(x)C by the first corollary above. So xf-1(C)=D too. Hence D is closed, again by the same corollary. ∎

Title properties of first countability
Canonical name PropertiesOfFirstCountability
Date of creation 2013-03-22 19:09:26
Last modified on 2013-03-22 19:09:26
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 11
Author CWoo (3771)
Entry type Result
Classification msc 54D99