If for every sequence $(x_n)$, $x_n \to x_0 \implies (f(x_n))$ is a Cauchy sequence, then is $f$ continuous?

Question

Given $f:D \subseteq \mathbb{R} \to \mathbb{R}$, if for every sequence $(x_n)$ in $D$ such that $x_n \to x_0$ is true that $(f(x_n))$ is a Cauchy sequence, then is $f$ continuous?

This is what I know:

1. Every Cauchy sequence is a convergent sequence.
2. $f$ is continuous if and only if $x_n \to x_0 \implies f(x_n) \to f(x_0)$ for every sequence $(x_n)$.

However, I only know that $f(x_n) \to a \in \mathbb{R}$; how can I see that $a = f(x_0)$ and then have the continuity of $f$?

If the result is false, please show a counterexample, i.e., a discontinuous function such that for every $x_n \to x_0$ I have $(f(x_n))$ Cauchy.

Answer 1

The statement is correct. To see this, let $x_0\in D$ and $(x_n)$ a sequence in $D$ with $x_n\to x_0$. We want to show that $f(x_n)\to f(x_0)$. Define the sequence $(y_n)=(x_1,x_0,x_2,x_0,x_3,x_0,\dots)$. Then $y_n\to x_0$, so by assumption $f(y_n)$ is a Cauchy-sequence. It has the constant convergent subsequence $f(x_0)$, hence $f(y_n)$ converges to $f(x_0)$ (A Cauchy-sequence with convergent subsequence is itself convergent with the same limit). As $f(x_n)$ is a subsequence of $f(y_n)$ it follows that $f(x_n)\to f(x_0)$

Answer 2

Correct me if wrong.

Perhaps a bit of nitpicking.

1)If $x_0 \in D;$

Then consider $x_n:=x_0$ and you are done

2)Assume $x_0\not \in D$, and $x_0$ a limit point of $D,$ then

$f(x_n) \rightarrow f(a);$ i.e. $f(x_n)$ has a limit;

Continuity does not follow.

Example:

$D_L=$ { $x|x<0$ };

$f:(x)=0$ for $x<0;$

Define $D'=$ { $0$ }:

$f(0)=1.$

$D_R$ defined similarly.