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In order to prove:

For $f:M\to N$ to be continuous its sufficient that $x_n\to a\implies f(x_n)_n$ is convergent in N

I'm supposing that $x_n$ is convergent, that is:

$$\forall \epsilon>0, \exists n_0 | n>n_0 \implies d(x_n, a)<\epsilon$$

Now, I must show that when all above implies $f(x_n)_n$ convergent, we have $f$ continuous. I must somehow prove that when te above implies $d(f(x_n), y)<\epsilon_2$ I have:

$$d(x,a)<\delta \implies d(f(x), f(a))<\epsilon$$

But how to relate $n$ and $x$? Or $x_n$ with $f(x)$?

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  • $\begingroup$ Do you mean that $f(x_n) \to f(a)$, or to any point? $\endgroup$ May 24, 2016 at 12:45
  • $\begingroup$ It just says 'convergent in N' $\endgroup$ May 24, 2016 at 12:54
  • $\begingroup$ It is well known that if $f(x_n) \to f(a)$ for all $x_n \to a$, $f$ is continuous. If you replace $f(x_n) \to f(a)$ by $f(x_n) \to y$ with $y \in N$ any point, I'm not sure. $\endgroup$ May 24, 2016 at 13:01
  • $\begingroup$ Since it is not specified that $x_n$ can't equal $x$, if there exists $(y_n)_n$ converging to $x$ with each $y_n\ne x,$ then let $x_{2 n}=y_n$ and $x_{2 n-1}=x. $ For $(f(x_n))_n$ to converge, it must converge to $f(x).$ $\endgroup$ May 24, 2016 at 13:09

2 Answers 2

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Let $(x_n)$ be a convergent sequence in $M$ with limit $a$. Then we can consider the sequence $x_1, a, x_2, a, x_3, a, \ldots$ which intertwines $(x_n)$ with the constant sequence $(a)$. This sequence converges to $a$ as well, so by assumption, the sequence $f(x_1), f(a), f(x_2), f(a), \ldots$ converges in $N$. Since every other term is $f(a)$, the only possible limit is $f(a)$ itself. Clearly, this implies that $f(x_n)$ converges to $f(a)$, so $f$ is indeed continuous.

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  • $\begingroup$ Nice. I was trying to find a counterexample, but clearly it's like this. $\endgroup$ May 24, 2016 at 13:10
  • $\begingroup$ What is the utility of the interwine sequence and why it converges to f(a)? $\endgroup$ May 24, 2016 at 13:24
  • $\begingroup$ @GuerlandoOCs Let $b$ be the limit of the sequence $f(x_1), f(a), f(x_2), f(a), \ldots$. Let $y_n$ denote the $n^{\text{th}}$ term of this sequence. Then for every $\epsilon > 0$, there is some $N$ so that for all $n>N$, $d(y_n, b) < \epsilon$. However we can always find an $n> N$ for which $y_n = f(a)$ (just choose $n$ to be an even number bigger than $N$). Thus, we conclude that $d(f(a), b) < \epsilon$ for every $\epsilon>0$. This can only be true if $d(f(a), b) = 0$, which implies that $b=f(a)$. That's the utility of thinking about the intertwined sequence. $\endgroup$
    – Alex G.
    May 24, 2016 at 13:28
  • $\begingroup$ @GuerlandoOCs (+1) This answer shows that if $f$ respects convergence in the sense of "$a_n$ convergent$\implies f(a_n)$ convergent", then it will also obey the rule "$a_n\to a\implies f(a_n)\to f(a)$". Having proved that, you can now finish the job by means of a counterexample as is given in the answer of user254665. $\endgroup$
    – drhab
    May 24, 2016 at 13:53
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By contradiction, suppose $f$ is discontinuous at $x.$ Then, from the $\delta$ -$\epsilon$ def'n of continuity, $$\exists \epsilon >0 \;\forall \delta >0\;\exists y\;(|x-y|<\delta \land |f(x)-f(y)|\geq \epsilon).$$ So take such an $\epsilon,$ and for each $n\in N,$ take $y_n$ such that $|x-y_n|<1/n$ and $|f(x)-f(y_n)|\geq \epsilon.$

Now for each $n\in N$ let $x_{2 n}=x$ and $x_{2 n-1}=y_n.$ Then $(x_n)_{n\in N}$ converges to $x$ but $(f(x_n))_{n\in N}$ does not converge to $f(x).$

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    $\begingroup$ Indeed, $f(y_n)_n$ does not converge to $f(x)$. But that does not exclude that $f(y_n)_n$ converges. $\endgroup$
    – drhab
    May 24, 2016 at 13:34
  • $\begingroup$ @drhab .Good point. I am editing my A. $\endgroup$ May 24, 2016 at 13:39

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