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Let $f\colon(X,d_X)\rightarrow(Y,d_Y)$ be a function between metric spaces such that for every convergent sequence $(x_n)_{n\in\mathbb{N}}$ in $X$ the sequence $(f(x_n))_{n\in\mathbb{N}}$ is convergent in $Y$. Does this impy continuity of $f$?

At first I thought that this does not imply the continuity of $f$, so I tried to think of a counterexample. I thought about it for a long time, but I couldn't find one. I found a few similar problems, but none helped me. Can someone help me please?

José Carlos Santos
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Tobi92sr
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    hint: If $f$ is not continuous at $x$, then there is some $\epsilon > 0$ such that for every $n > 0$ there exists some $x_n$ with $d(x, x_n) < \frac 1n$ and $d(f(x), f(x_n)) > \epsilon$ – Paul Sinclair Oct 18 '17 at 16:18
  • Do you know the difference between the "limit of the function at some point $x_0$" and "the function is continuous in $x_0$"? There is a very subtle one, with the latter, the function must be defined in $x_0$, i.e. $f(x_0)$ must exist. With the definition of the limit, that's not required. – rtybase Oct 18 '17 at 16:37

2 Answers2

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Suppose that $f$ is discontinuous at some point $x$. Then,for some $\varepsilon>0$, if $n\in\mathbb N$, then there is a $x_n\in B\left(x,\frac1n\right)$ such that $d\bigl(f(x),f(x_n)\bigr)\geqslant\varepsilon$. Now consider the sequence$$x,x_1,x,x_2,x,x_3,\ldots$$It converges (to $x$). However, the sequence$$f(x),f(x_1),f(x),f(x_2),f(x),f(x_3),\ldots$$does not converge.

José Carlos Santos
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Assume $f$ is not continuous at $x \in X$. Then there must exist $\{x_n\} \rightarrow x$, and so $f(x_n)$ is convergent in $Y$, but $f(x_n) \nrightarrow f(x)$. Now if $x$ was an isolated point of $X$ then $x_n = x \forall n > N, N \in \Bbb N$ and it's obvious $f$ must be continuous at $x$ (think why).

Otherwise, if $\{x_n\}$ isn't that trivial, then we can consider the following sequence:

$$ y_{2n} = x, y_{2n+1} = x_n \implies y_n \rightarrow x $$

and, of course, $\{f(y_n)\}$ converges, too.

But then $f(x) = \lim_{n \to \infty} f(y_{2n}) = \lim_{n \to \infty} f(y_{2n+1}) = \lim_{n \to \infty} f(x_n)$,

and we know this is equivalent to $f$ being continuous at $x$ in metric spaces.

cronos2
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    If $f(x)$ is not defined why do we even bother checking whether $f$ is continuous or not at $x$ (even if that had a meaning) – cronos2 Oct 18 '17 at 16:53
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    But for $f$ to be not continuous there must exist some $x$ in the domain (where the continuity condition fails), and thus $f(x)$ is defined, unless the domain is empty which would make for a rather stupid question. – cronos2 Oct 18 '17 at 17:02
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    @rtybase We are given that $f$ is a function defined on $X$. Thus $f(x)$ exists for every $x\in X$. – John Griffin Oct 18 '17 at 17:03