We say that a function, $f: X \to Y$ ($X, Y$ are topological spaces) is sequentially continuous if $f(x_n)$ converges to $f(x)$ whenever $x_n$ converges $x$. Give an example of a function that is sequentially continuous but not continuous.

I tried letting $X$ be $\mathbb R$ with cofinite topology and $Y$ be $\mathbb R$ with discrete topology where $f$ is identity map but $f$ is neither sequential continuous or continuous. There are no simple solutions online since most use ordinal sets and we have not yet covered that in our topology class.

  • 1
    $\begingroup$ This is not really a duplicate because it specifically asks for an answer that does not use ordinals (or other such advanced machinery). $\endgroup$ Commented Sep 30, 2015 at 1:11
  • $\begingroup$ I agree.The other Q has 3 answers ,2 of which use ordinals and the other says to take the characteristic function of a subset A of a space X, where the closure of A is not equal tp its sequential closure. Which works, but then you have to find such A and X, which is basically the same Q. (I didn't vote on it.) $\endgroup$ Commented Sep 30, 2015 at 16:19
  • $\begingroup$ Tiny typo in the Q :"converges x" should be "converges to x". $\endgroup$ Commented Sep 30, 2015 at 16:20

1 Answer 1


Note that in the following, by countable I mean not uncountable, that is, countable means"finite or countably infinite".

Consider a space $X$ in which every non-empty $V\subset X$ is open if and only if $X\backslash V$ is countable .This is called the co-countable topology on $X$. A sequence $(p_n)_{n \in \mathbb N}$ of points in $X$ cannot converge, in any sense, to a point $p \in X$ if $\{n \in\mathbb N : p_n=p\}$ is finite. Hence any convergent sequence in $X$ is eventually constant, and therefore any function $f:X\to Y$ to any space $Y$ is sequentially continuous. Now if $X$ is uncountable then it is not a discrete space so there exist discontinuous functions on $X$. For an example, let $X=\mathbb R$ (the reals) and let $Y$ be the reals with the usual topology, and let $f=\text{id}_{\mathbb R}$. The inverse $f^{-1}(0,1)$ of the open interval $(0,1)$ is not open in $X$. For another example, let $X$ be any uncountable set, with the co-countable topology, let $Y$ be the set $X$ with the discrete topology, and let $f=\text{id}_X$. Then $\{p\}$ is open in $Y$ for any $p \in Y$ but $f^{-1}\{p\}=\{p\}$ is not open in $X$.

  • $\begingroup$ Your last case doesn't work since it's not sequentially continuous. $\endgroup$
    – Yunus Syed
    Commented Sep 30, 2015 at 1:18
  • 1
    $\begingroup$ Why not? On the co-countable topology any function is sequentially continuous because convergent sequences are eventually constant. $\endgroup$ Commented Sep 30, 2015 at 6:53
  • $\begingroup$ Oh sorry I misunderstood co-countable as finite complement. $\endgroup$
    – Yunus Syed
    Commented Sep 30, 2015 at 12:46
  • $\begingroup$ Some people use "countable" to mean countably infinite. Some don't. This make it easy to make a slip. $\endgroup$ Commented Sep 30, 2015 at 15:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .