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 $f(x)$.
[Math] Example of topological spaces where sequential continuity does not imply continuity
continuityexamples-counterexamplesgeneral-topology
Related Solutions
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 $X$.
So called "sequential continuity" does not characterize continuity. As the theorem says in Munkres, you only get the converse when $X$ is first countable.
For a counterexample, see the following
Sequentially continuous but not continuous
You can characterize continuity using nets instead of sequences though. An explanation of this is also in Munkres, although it is relegated to the exercises.
The other statement you want to prove is true though.
Let $f:X\rightarrow Y$ be a function and $p\in X$ a point admitting a countable neighbourhood basis $\{U_{n}\}_{n\in\mathbb{N}}$. Then $f$ is continuous at $p$ if and only if for every sequence $(x_{n})$ in $X$ that converges to $p$ we have that $(f(x_{n}))$ converges to $f(p)$.
We may assume that the neighbourhood basis of $p$ is such that $n<m$ implies that $U_{n}\subseteq U_{m}$.
Assume that $f$ satisfies the sequential condition at $p$. We then let $V\subseteq Y$ be an open neighbourhood of $f(p)$. Assume towards a contradiction that for all $n$ we have that $f(U_{n})\not\subseteq V$. Then for each $n\in\mathbb{N}$ we may choose an $x_{n}\in U_{n}$ such that $f(x_{n})\notin V$. However, the sequence $(x_{n})$ converges to $p$ (easily checked). Then $(f(x_{n}))$ must converge to $f(p)$, but as $f(x_{n})\notin V$ for all $n$, this is impossible. This is a contradiction. Therefore, there is some $n\in\mathbb{N}$ for which $f(U_{n})\subseteq V$, establishing continuity at $p$.
Now assume that $f$ is continuous at $p$ in the sense that for all open $V\subseteq Y$ containing $f(p)$ there is some open $U\subseteq X$ such that $p\in U$ and $f(U)\subseteq V$. We then let $(x_{n})$ be a sequence in $X$ converging to $p$. We need to show that $(f(x_{n}))$ converges to $f(p)$. If$(f(x_{n}))$ does not converge to $f(p)$ there is an open neighbourhood $V\subseteq Y$ of $f(p)$ that doesn't contain infinitely many of the points $f(x_{n})$. By assumption there is an open $U\subseteq X$ containing $p$ such that $f(U)\subseteq V$. Because $(x_{n})$ converges to $p$ we have that there is an $N\in\mathbb{N}$ such that for all $m\geq N$ we have that $x_{m}\in U$. That is, there are only finitely many $x_{n}$ that are not in $U$. Because $f(U)\subseteq V$ we have that there are only finitely many $f(x_{n})$ not in $V$, contradicting there being infinitely many $f(x_{n})$ not in $V$. Therefore we must have that $(f(x_{n}))$ converges to $f(p)$.
Edit: I'm aware that the image of the sequence $(x_{n})$ may be finite. When I say "$V$ doesn't contain infinitely many of the points $f(x_{n})$" I mean that there is an infinite subset $K\subseteq\mathbb{N}$ such that $f(x_{s})\notin V$ for all $s\in K$.
Best Answer
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 $X$.