Our class's lecture notes approached it indirectly so I want my worked to be checked here.

Proposition: Let $f:(X,\rho)\rightarrow (Y,\sigma)$ be a continuous function between two metric spaces. Then $f$ is continuous at $x\in X$ if and only if for every sequence $(x_n)$ that converges to $x$, then the corresponding sequence $(f(x_n))$ converges to $f(x)$.

Proof attempt:

Suppose $\sigma(f(x_n),f(x))<\epsilon$. Since $f$ is continuous at $x\in X$ then $\exists \delta >0$ such that for all $\alpha\in X$, we have that if $\rho(x,\alpha)<\delta$, then $\sigma(f(x),f(\alpha))<\epsilon$. But since $(x_n)$ is convergent to $x$, then there exists $N\in\Bbb{N}$ such that $\forall n\geq N$, $\rho(x,x_n)<\delta$. Thus, for all $n\geq N$, $\sigma(f(x_n),f(x))<\epsilon$. Hence, $(f(x_n))$ converges to $f(x)$. Conversely, if $(f(x_n))$ converges to $f(x)$ then $\forall\epsilon'>0\exists M\in\Bbb{N}$ such that $\forall m\geq M$, $\sigma(f(x_m),f(x))<\epsilon'$. Note that $\forall m\geq M$, $\rho(x,x_m)<\delta'$, for some $\delta'>0$. Since $(x_n)$ is convergent to $x$, $\exists M_1\in\Bbb{N}$ so that for all $i\geq M_1$, $\rho(x,x_i)<\delta'$. Therefore, $f$ is continuous at $x$.


The converse (if the sequence condition holds at $x$, then $f$ is continuous at $x$) has to be an indirect proof. The forward direction you almost got right, except that the first sentence is nonsense: you assume $x_n \to x$ and you know continuity $\epsilon-\delta$ definition) and you need to show that $f(x_n) \to f(x)$, so to show the last with the definition of convergence in $Y$ you have to start by picking $\epsilon>0$.

By continuity of $f$ at $x$, there exists a $\delta>0$ such that for all $x' \in X$: if $d(x,x') < \delta$ we know that $d(f(x'), f(x)) < \epsilon$. So in order to get the $f(x_n)$ $\epsilon$-close to $f(x)$ we know that $x_n$ has to be $\delta$-close to $x$, and this will happen eventually: there exists $N \in \mathbb{N}$ such that for all $n \ge N$ we have that $d(x_n, x) < \delta$. But then for this same $N$ we thus have that $d(f(x_n), f(x)) < \epsilon$ for all $n \ge N$, as required. You can see how the definition of convergence nicely interacts with the continuity: both are about eventual approximation.

For the reverse we have to show that for each $\varepsilon > 0$ there exists some $\delta >0$ such that for all $x'$ $\delta$-close to $x$, $f(x')$ is $\varepsilon$-close to $f(x)$. And all are given is a statement about convergent sequences. Now going for a contradiction we at least get a positive statement we can work with: if $f$ is not continuous at $x$, there exists some fixed $\epsilon >0$ such that no $\delta>0$ will "work", so for all $\delta>0$, there is some $x(\delta)$ (the point depends on how we choose the $\delta$, hence the notation) such that $d(x(\delta), x) < \delta$ but still $d(f(x(\delta)), f(x)) \ge \epsilon$.

Now a standard way to make a sequence in $X$ that converges to $x$, is to pick a point $x_n$ with $d(x_n, x)< \frac1n$ (or some other term in $n$ that goes to $0$ for larger and larger $n$). We can do this by our contradiction assumption: let $x_n = x(\frac1n)$ in the above. Then we know that $x_n \to x$ so now the sequence assumption can finally be used! We conclude that $f(x_n) \to f(x)$ but his is a contardiction, as in that case $f(x_n)$ should get $\epsilon$-close to $f(x)$ for large enough $n$, while for all $n$ these are at least $\epsilon$ apart, contradiction. So $f$ is continuous at $x$.

So you should have to product a $\delta>0$ for this proof of continuity. You cannot do that easily, but if you assume no such $\delta$ exists, we can construct a sequence on which to apply the sequence assumption. The indirectness seems quite unavoidable to me.


For the first you should delete the first sentence. The rest of this part is OK. The converse part is completely wrong. What sequence $\{x_n\}$ are you using to prove continuity of $f$ at $x$? An arbitrary sequence converging to $x$ will not lead to the conclusion. For a correct proof use proof by contradiction. Suppose $f$ is not continuous at $x$. There exists $\epsilon >0$ such that for any $\delta >0$ there exists $y$ with $d(x,y)<\delta $ but $\sigma (f(y),f(x)) >\epsilon $. Apply this with $\delta =1,\frac 1 2, \frac 1 3,...$. You will get points $y_n,n=1,2,...$. Show that this sequence converges to $x$ but $\{f(y_n)\}$ does not converges to $f(x)$.

  • $\begingroup$ I'm confused. For the converse part,the assumption is for all sequence $(x_n)$ that converge to $x$, the corresponding sequence $(f(x_n))$ converges to $f(x)$. So why can't I use an arbitrary sequence $(x_n)$? $\endgroup$ – TheLast Cipher Jul 12 '18 at 8:34
  • $\begingroup$ What is the definition of continuity at $x$ you are using? $\endgroup$ – Kavi Rama Murthy Jul 12 '18 at 8:38
  • $\begingroup$ $f$ is continuous at $x$ iff for all $\epsilon>0\exists\delta>0$ such that if $\rho(x,y)<\delta$ then $\sigma(f(x),f(y))<\epsilon$ $\endgroup$ – TheLast Cipher Jul 12 '18 at 8:45
  • $\begingroup$ So you are not given any sequence $\{x_n\}$ in this definition. The important point is the $\delta $ you get in the proof cannot depend on some sequence $\{x_n\}$. You should start with $\epsilon $ and produce a $\delta $ in terms of it. Is it clear now that you cannot start the proof of continuity by taking some arbitrary sequence $\{x_n\}$ converging to $x$?. $\endgroup$ – Kavi Rama Murthy Jul 12 '18 at 9:08
  • $\begingroup$ I understood from earlier lessons without metric spaces that the terms cannot depend on $\delta$, but I am still trying to determine why you see that the terms of the arbitrary sequence $(x_n)$ depended on $\delta'$ since the way I see it is that I merely pointed out that the terms of the sequence $(x_n)$ will eventually be within $\delta'$ of $x$. $\endgroup$ – TheLast Cipher Jul 12 '18 at 9:26

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