Let $X,Y$ be topological spaces and $f: X \to Y$. I know that if $X,Y$ are not necessarily first countable (=countable nbhood base) then
''For all sequences $x_n\to x$ in $X$ it's true that $f(x_n) \to f(x)$ in $Y$''
does not imply that $f$ is continuous.
I am trying to find an example of $f,X,Y$ such that for all sequences $x_n\to x$ in $X$ it's true that $f(x_n) \to f(x)$ in $Y$ but $f$ is discontinuous.
My first idea is $X=\mathbb R$ with the usual topology and $Y=\mathbb R$ with the discrete topology and then $f(x) = x$. Then $f$ is not continuous but also $f(x_n) \not\to f(x)$ even if $x_n \to x$. I am trying to think of spaces that do not have countable nhood bases but all I can think of is the discrete topology in an uncountable set.
Could anybody help me with an example?
One of the simplest examples is using $X = \mathbb{R}$ in the co-countable topology (i.e. the closed sets are $X$ and the countable (including finite and empty) sets), and $Y$ as $\mathbb{R}$ in the usual topology, and as a map $f(x) = x$, the identity.
This function is not continuous (an open interval in the usual topology is not open in co-countable topology, as its complement is not countable), but $X$ has the property that any sequence $(x_n)$ in it that converges to some $x$ is eventually constant, i.e. for some $N$, all $x_n$ with $n > n$ are equal to $x$.
This follows from considering $A = \{x_n: x_n \neq x\}$ and noting that $O = X\setminus A$ is an open neighbourhood of $x$ and so all terms must be eventually in $O$....
So for any convergent sequence $(x_n)$, the image sequence under $f$, i.e. the same sequence but considered in $Y$, is also convergent (as eventually constant sequences are convergent to that constant in any topology, by definition). So $f$ is (trivially) sequentially continuous but as said, not continuous.