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In $\mathbb R$, all Cauchy sequences are convergent and all convergent sequences are Cauchy. So, why isn't continuity enough to preserve Cauchy sequences?

A function is continuous iff it preserves convergent sequences.

A sequence is convergent iff it is Cauchy.

So, why doesn't it follow that continuous functions preserve Cauchy sequences?

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  • $\begingroup$ Continuity of what, you meant $(a_n)$ is Cauchy then so is $(f(a_n))$ ? What if $f$ is continuous at $l = \lim_n a_n$ ? $\endgroup$ – reuns Nov 6 '17 at 3:10
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    $\begingroup$ Think of $f(x) = 1/x$ on $(0,1)$ and $a_n = 1/n$. $\endgroup$ – amsmath Nov 6 '17 at 3:18
  • $\begingroup$ Ok, we have $a_n \to 0$, $f(a_n)=1/a_n=n$. So, $f(a_n) \to \infty$. What does this tell me? $\endgroup$ – Al Jebr Nov 6 '17 at 3:38
  • $\begingroup$ "All Cauchy sequences are convergent" False. $\endgroup$ – zhw. Nov 6 '17 at 3:42
  • $\begingroup$ Ok, I'm considering $\mathbb R$ only. $\endgroup$ – Al Jebr Nov 6 '17 at 3:44
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why isn't continuity enough to preserve Cauchy sequences?

Because, in general, it is not given that the limit of the sequence belongs to the domain of the function.

The implication $$\lim_{n\to\infty} x_n= x\quad \Longrightarrow\quad \lim_{n\to\infty} f(x_n)= f(x)$$ requires $x_n,x\in D(f)$. Thus, if $x\notin D(f)$, this argument does not work (a counterexample was given in the comments of your post). In fact, in this case we need an extra condition (see the first comment below).


Edit

Let us analyze your argument:

Claim: Continuous functions preserve Cauchy sequences.

Poof: Let $X$ be a subset of $\mathbb R$. Let $f:X\to\mathbb R$ be a continuous function. Let $(x_n)$ be a Cauchy sequence.

We want to show that $(f(x_n))$ is Cauchy.

  1. As a (real) sequence is convergent iff it is Cauchy, there exists $x_0\in\mathbb R$ such that $$\lim_{n\to\infty}x_n=x_0.\tag{$*$}$$

  2. As a function is continuous iff it preserves convergent sequences, there exists $L\in\mathbb R$ such that $$\lim_{n\to\infty}f(x_n)=L.$$

  3. As a (real) sequence is convergent iff it is Cauchy, we conclude that $(f(x_n))$ is Cauchy.

The problem is step 2 which, in general, does not follow from $(*)$. If $x_0$ belongs to $X$, then it is true (with $L=f(x_0)$) but in general we cannot guarantee the existence of $L$.

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    $\begingroup$ (we need continuity on the closed set $\overline{D(f)}$, for $D(f)$ bounded it is the same as uniform continuity) $\endgroup$ – reuns Nov 6 '17 at 3:34
  • $\begingroup$ @reuns Thanks, I edited my post. $\endgroup$ – Pedro Nov 6 '17 at 3:40
  • $\begingroup$ But what is wrong with the implications: In $\mathbb R$, $x_n$ Cauchy implies $x_n$ convergent implies $f(x_n)$ convergent implies $f(x_n) Cauchy? Doesn't this mean continuous functions preserve Cauchy? $\endgroup$ – Al Jebr Nov 6 '17 at 3:48
  • $\begingroup$ @AlJebr Nothing. See the Additional comment here. $\endgroup$ – Pedro Nov 6 '17 at 3:50
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    $\begingroup$ @AlJebr I must say "for every sequence $a_n \to l$, $f(a_n) \to f(l)$" is the very definition of "$f$ is continuous at $l$". Now $f(x) = 1/x, x \in (0,1]$ is continuous but not uniformly continuous on $(0,1]$, thus it is not continuous at $0$ and $\lim_{n \to \infty} f(1/n)$ diverges. $\endgroup$ – reuns Nov 6 '17 at 3:53

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