Nets and the Axiom of Choice - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:26:28Z http://mathoverflow.net/feeds/question/50371 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50371/nets-and-the-axiom-of-choice Nets and the Axiom of Choice mathahada 2010-12-25T20:31:22Z 2010-12-25T21:21:57Z <p>Suppose that $f : X \rightarrow Y$ is mapping between topological spaces that is not continuous at $x_0$. Then there is an open set $V$ in $Y$ containing $f(x_0)$ such that for any open set $U$ containing $x_0$, there is some $x_U \in U$ with $f(x_U) \notin V$. By picking one from each $U$ we can build a net $x_U$ converging to $x_0$ such that $f(x_U)$ does not converge to $f(x_0)$. This however requires the axiom of choice because we have to pick $x_U$ from each $U$. </p> <p>My question is: if we are given that a function between topological spaces is continuous if and only if it is "sequentially" continuous (in the sense of nets, not necessarily infinite sequences indexed by the integers) - does the axiom of choice follow?</p> http://mathoverflow.net/questions/50371/nets-and-the-axiom-of-choice/50377#50377 Answer by Bill Johnson for Nets and the Axiom of Choice Bill Johnson 2010-12-25T21:21:57Z 2010-12-25T21:21:57Z <p>The axiom of choice is not needed if $X$ is $T_1$; i.e., singleton sets are closed.</p> <p>Suppose $f:X\to Y$ is not continuous at $x_0$. Let $D$ be the collection of all $(U,x)$ s.t. $U$ is an open set containing both $x_0$ and $x$ and $x_0\not=x$. Say that $(U,x) \le (V,y)$ provided $V\subset U$ and $x$ is not in $V$. If $X$ is $T_1$, this is a directed set. Define a net $g$ with domaine $D$ by $g(U,x)=x$. Then $g$ converges to $x_0$ but $f(g)$ does not converge to $f(x_0)$.</p>