2

Let $f:X\rightarrow Y$ where $X$ and $Y$ are topological spaces. Is it true that $f$ is continuous at a point $c$ iff for every sequence $x_n$ converging to $c$, we have $f(x_n)$ converging to $f(c)$?

idpd15
  • 1,994
  • 1
  • 17
  • 38
  • Top, spaces are a very broad class and includes spaces in which no sequence $(p_n)_{n\in N}$ converges to $p$ unless it is "eventually constant" (i,e, $ p_n=p$ for all but finitely many $n.$) – DanielWainfleet Jul 12 '16 at 16:01

2 Answers2

4

No. This holds if an only if $X$ is a sequential space. You can find counterexamples in the example section of that article.

joriki
  • 238,052
  • Thanks! I came across this counterexample but I didn't understand it. http://math.stackexchange.com/a/1419972/127598 . He says that convergence in X means that the sequence is eventually constant. How? – idpd15 Jul 12 '16 at 10:32
  • @idpd15: Convergence to a limit $x$ means that for any open neighbourhood of $x$ the sequence is eventually in that neighbourhood. Consider the complement $U$ of the set of elements of the sequence $\ne x$. As it's the complement of a countable set and contains $x$, this set is an open neighbourhood of $x$. The sequence is eventually in $U$, and since $U$ doesn't contain any elements of the sequence other than $x$, that's only possible if the sequence is eventually constant at $x$. – joriki Jul 12 '16 at 13:38
0

The topological generalization of the "$\delta, \epsilon $" def'n of continuity at $x$ of a function from $R$ to $R,$ is :

For $f:A\to B$ and $x\in A,$ we say $f$ is continuous at $x$ iff whenever $f(x)\in V\subset B$ and $V$ is open in $B,$ there exists $U\subset A$ where $p\in U,$ and $U$ is open in $A,$ and $f(q)\in V$ for all $q\in U.$

The topological version of $\lim_{n\to \infty }p_n=p$ is $$\cap_{n\in N} Cl (\{p_j: j\geq n\})=\{p\}.$$ Example:

Let $S$ be an infinite set. For $T\subset S$ let $T$ be closed iff $T$ is finite or $T=S.$ So $U\subset S$ is open iff $S \backslash U$ is finite or $U=\emptyset.$ Now if $V$ is any infinite subset of $S$ then $$Cl(V)=S.$$ Because otherwise $S \backslash Cl(V)$ is a non-empty open set that is disjoint from the infinite set $V,$ which is not possible.

Now let $(p_n)_{n\in N}$ be a sequence in $S.$ Consider any $p\in S.$

(1). If $\{p_n: n\in N\}$ is infinite, then $\{p_j: j\geq n\}$ is infinite for every $n\in N,$ so $\cap_{n\in N}Cl(\{p_j:j\geq n\}=$ $\cap_{n\in N}S=S,$ and $(p_n)_{n\in N}$ does not converge to any member of $S.$

(2).If $\{p_n: n\in N\}$ is finite but $\{n:p_n\ne p\}$ is infinite then there exists $q\ne p$ such that $\{n:p_n=q\}$ is infinite, so $q\in \{p_j: j\geq n\}$ for every $n$. So $q\in \cap_{n\in N}\{p_j:j\geq n\}$ and $(p_n)_{n\in N}$ does not converge to $p.$

Therefore $(p_n)_{n\in N}$ does not converge to $p$ unless $\{n:p_n\ne p\}$ is finite.

DanielWainfleet
  • 57,985