Sequential Continuity does not imply Continuity

Question

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.

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 $X$.

Answer 2

Let $S=\{0,1\}$, the Sierpiński topology on $S$ is the collection $\{\emptyset,\{1\},\{0,1\} \}$.

Let $X$ be a topological space, $M$ be a sequentially closed subset of $X$ and $M$ is not closed, then the function$$f:X\to S,f^{-1}\{0\}=M$$is sequentially continuous and $f$ is not continuous.

So if we have an example of sequentially closed but not closed set, then we can get an example of sequentially continuous but not continuous function.

The following is an example of a sequentially closed but not closed set.

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Let $X=\ell^2$ and let $M=\left\{n^{1 / 2} e_n: n \geq 1\right\} \subseteq X$ with weak topology. We will show that $0\in\overline M$ but that there exists no sequence $\left(x_n\right)$ with terms in $M$ such that $x_n \to 0$ weakly as $n \to \infty$.

Let $U=\left\{x \in \ell^2:\left|\left(x, y_k\right)\right|<\varepsilon\right.$ for $\left.1 \leq k \leq K\right\}$ be a basic weakly open neighbourhood of $0$. If $M \cap U=\emptyset$ then for every $n \geq 1$ there exists $k_n \in\{1, \ldots, K\}$ such that $\left|\left(n^{1 / 2} e_n, y_{k_n}\right)\right| \geq \varepsilon$. Thus $\left|y_{k_n}(n)\right| \geq \varepsilon n^{-1 / 2}, n \geq 1$, and hence $$ \sum_{k=1}^K\left\|y_k\right\|_2^2 \geq \sum_{k=1}^K \sum_{n=1}^N\left|y_k(n)\right|^2=\sum_{n=1}^N \sum_{k=1}^K\left|y_k(n)\right|^2 \geq \varepsilon^2 \sum_{n=1}^N \frac{1}{n}, \quad N \geq 1 $$ which is absurd because the left-hand side is finite and the right-hand side is unbounded as $N \rightarrow \infty$. It follows that $0$ lies in the weak closure of $M$.

Suppose that $(x_n)$ is a sequence in $M$ which converges weakly to some limit. Then by a standard application of the Uniform Boundedness Principle the sequence $(x_n)$ is norm-bounded and hence must be eventually constant. Since $0 \notin M$ there exists no sequence in $M$ which converges weakly to $0$.