I have anexo exercise to answer that if for every sequence $x_{n} \rightarrow x$ we have $f(x_{n}) \rightarrow f(x)$ than is f continuous or not. Continuous here means that $f^-1(A)$ is open If $A$ is open I feel that this is false but can neither proof or show a counter example PS: Sorry If duplicate but couldnt find the same question on search
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Do you know about uncountable ordinals? – Angina Seng Jan 09 '19 at 19:53
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@LordSharktheUnknown never heard of but can look It up – Daniel Moraes Jan 09 '19 at 20:00
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One can only guarantee that a function $f:X \to Y$ is continuous $\iff$ for every sequence $x_n$ in $X$ converging to $x \in X$ we have that $f(x_n)$ converges to $f(x)$ if $X$ is a so called $\textit{first countable}$ space. If $X$ is not first countable, then one cannot guarantee that the latter implies the former. A first countable space is a topological space $X$ where every $x \in X$ has a countable neighbourhood basis. In the comment by $\textit{Cheerful Parsnip}$ there is a link to a nice counterexample showing this.
greelious
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