Example of topological spaces where sequential continuity does not imply continuity

Please give an example of a function $f : X \to Y $ where $X,Y$ are topological space , such that there exist $x \in X$ such that for every sequence $\{x_n\}$ in $X$ converging to $x$ , $\{f(x_n)\}$ converges to $f(x)$ but $f$ is not continuous at $x$ ; also please give such an example that $f$ is not continuous any where in the domain but for every $x \in X$ and sequence $\{x_n\}$ in $X$ converging to $x$ , $\{f(x_n)\}$ converges to $f(x)$.

Solution 1:

Let $X=(\Bbb R,\tau_{cc})$ be the real line with the cocountable topology, i.e. closed sets are the countable sets in $\Bbb R$. Note that any subset $A$ of $X$ is sequentially closed since $A$ contains the limit of every convergent sequence in $A$, as convergence in $X$ means that a sequence is eventually constant.
Let $Y$ be the discrete real line, and let $f:X\to Y$ be given by the identity. Clearly $f$ is sequentially continuous, however, it is not continuous at any point $x$, since continuity at $x$ means that $\{x\}$ is open in $X$.

Solution 2:

Here is an example where the function $f$ is everywhere sequentially continuous but nowhere continuous, the space $X$ is completely regular (in particular Hausdorff) and the space $Y$ is finite discrete.

Let $I$ be an uncountable set and consider the space $2^I$ with its product topology, considered as the set of all functions $x : I \to \{0,1\}$. Note that $2^I$ is compact Hausdorff by Tychonoff's theorem; in particular $2^I$ is completely regular.

Let $X_0$ be the set of all $x \in 2^I$ such that $x^{-1}(\{1\})$ is countable (so an element of $X_0$ "mostly" takes the value 0). By definition of the product topology, $X_0$ is dense in $2^I$. Also, $X_0$ is sequentially closed. To see this, suppose $x_1, x_2, \dots \in X_0$ and $x_n \to x$, so that $x_n(i) \to x(i)$ for every $i \in I$. If we let $A_n = x_n^{-1}(\{1\})$, which is countable, then it's clear that $x^{-1}(\{1\}) \subset \bigcup_n A_n$ which is thus also countable. So $x \in X_0$.

Likewise let $X_1$ be the set of all $x \in 2^I$ such that $x^{-1}(\{0\})$ is countable. Then $X_1$ is also dense and sequentially closed, and $X_0 \cap X_1 = \emptyset$.

Set $X = X_0 \cup X_1$ with the subspace topology inherited from $2^I$. Then $X$ is also completely regular.
Let $Y = \{0,1\}$ with the discrete topology, and define $f : X \to Y$ by $$f(x) = \begin{cases} 0, & x \in X_0 \\ 1, & x \in X_1. \end{cases}$$ Now $f$ is sequentially continuous everywhere, because $X_0, X_1$ are sequentially closed. But for any nonempty $U \subset X$, since $X_0, X_1$ are both dense in $X$, we see that $f$ takes both the values $0$ and $1$ on $U$. Therefore $f$ is nowhere continuous.

For (locally) compact Hausdorff examples, see A function on an LCH space that is sequentially continuous but nowhere continuous.

Solution 3:

Let $\omega_1$ be the first uncountable ordinal and take $X=[0, \omega_1]$ (so really $\omega_1\cup\{\omega_1\}$) with the order topology. Every sequence converging in $[0,\omega_1)$ is bounded, hence every sequence in $X$ which converges to $\omega_1$ must be eventually constant. Therefore any $f\colon X\to Y$ is sequentially continuous in $\omega_1$. Now you set take $f(x)=0$, if $x<\omega_1$ and $f(\omega_1)=1$. This function is not continuous in $\omega_1$, because $\{\omega_1\}$ is not open, since $\omega_1$ is not a successor.