testing for continuity via nets

Proposition 1.

Let X,Y be topological spacesMathworldPlanetmath and f:XY. Then the following are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath:

  1. 1.

    f is continuousMathworldPlanetmathPlanetmath;

  2. 2.

    If (xi) is a net in X converging to x, then (f(xi)) is a net in Y converging to f(x).

  3. 3.

    Whenever two nets (xi) and (yj) in X convergePlanetmathPlanetmath to the same point, then (f(xi)) and (f(yj)) converge to the same point in Y.


(1)(2). Let A be the (directed) index setMathworldPlanetmathPlanetmath for i. Suppose f(x)U is open in Y. Then xf-1(U) is open in X since f is continuous. By assumptionPlanetmathPlanetmath, (xi) is a net, so there is bA such that xjf-1(U) for all jb. This means that f(xj)U for all ib, so (f(xi)) is a net too.

Conversely, suppose f is not continuous, say, at a point xX. Then there is an open set V containing f(x) such that f-1(V) does not contain any open set containing x. Let A be the set of all open sets containing x. Then under reverse inclusion, A is a directed setMathworldPlanetmath (if U1,U2A, then U1U2A). Define a relationMathworldPlanetmath RA×X as follows:

(U,x)R  iff  xU-f-1(V).

Then for each UA, there is an xX such that (U,x)R, since Uf-1(V). By the axiom of choiceMathworldPlanetmath, we get a function dR from A to X. Write d(U):=xU. Since A is directed, (xU) is a net. In addition, (xU) converges to x (just pick any UA, then for any WU, we have xW by the definition of A). However, (f(xU)) does not converge to f(x), since xUf-1(V) for any UA.

(2)(3). Suppose nets (xi) and (yj) both converge to zX. Then, by assumption, (f(xi)) and (f(yj)) are nets converging to f(z)Y.

Conversely, suppose (xi) converges to x, and i is indexed by a directed set A. Define a net (yi) such that yi=x for all iA. Then (yi)=(x) clearly converges to x. Hence both (f(xi)) and (f(yi)) converge to the same point in Y. But (f(yi))=(f(x)) converges to f(x), we see that (f(xi)) converges to f(x) as well. ∎

Remark. In particular, if X,Y are first countable, we may replace nets by sequences in the propositionPlanetmathPlanetmath. In other words, f is continuous iff it preserves converging sequences.

Title testing for continuity via nets
Canonical name TestingForContinuityViaNets
Date of creation 2013-03-22 19:08:58
Last modified on 2013-03-22 19:08:58
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 7
Author CWoo (3771)
Entry type Result
Classification msc 54C05
Classification msc 26A15
Related topic Net
Related topic FirstCountable