10

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?

user5826
  • 12,524
  • Continuity of what, you meant $(a_n)$ is Cauchy then so is $(f(a_n))$ ? What if $f$ is continuous at $l = \lim_n a_n$ ? – reuns Nov 06 '17 at 03:10
  • 2
    Think of $f(x) = 1/x$ on $(0,1)$ and $a_n = 1/n$. – amsmath Nov 06 '17 at 03:18
  • 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? – user5826 Nov 06 '17 at 03:38
  • "All Cauchy sequences are convergent" False. – zhw. Nov 06 '17 at 03:42
  • Ok, I'm considering $\mathbb R$ only. – user5826 Nov 06 '17 at 03:44
  • @zhw. All cauchy sequences are convergent in $\mathbb R$. – user5826 Nov 06 '17 at 03:45
  • @AlJebr With respect to the example, we have $f(a_n)-f(a_m)=n-m$. This is the point. – Pedro Nov 06 '17 at 03:47
  • @AlJebr Yes of course. But who said we were in $\mathbb R?$ How about continuous functions on $(0,1)?$ – zhw. Nov 06 '17 at 03:55
  • Is this correct? Suppose $f$ is continuous on $(0,1)$. So, if $x_n$ is a sequence in $(0,1)$, converging to $x_0 \in (0,1)$, then $f(x_n) \to f(x_0)$. Since $x_n \to x_0$, then $x_n$ is a Cauchy sequence. Also since $f(x_n) \to f(x_0)$ then $f(x_n)$ is a Cauchy sequence. – user5826 Nov 06 '17 at 04:14

1 Answers1

8

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 $L$.

Pedro
  • 19,965
  • 9
  • 70
  • 138
  • 2
    (we need continuity on the closed set $\overline{D(f)}$, for $D(f)$ bounded it is the same as uniform continuity) – reuns Nov 06 '17 at 03:34
  • @reuns Thanks, I edited my post. – Pedro Nov 06 '17 at 03:40
  • 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? – user5826 Nov 06 '17 at 03:48
  • @AlJebr Nothing. See the Additional comment here. – Pedro Nov 06 '17 at 03:50
  • @AlJebr There is nothing wrong if the domain is $\mathbb R$. – Pedro Nov 06 '17 at 03:53
  • 1
    @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. – reuns Nov 06 '17 at 03:53
  • I don't understand. If $0$ is not in the domain, why are we considering it? – user5826 Nov 06 '17 at 04:02
  • @AlJebr Because it is the limit of the given Cauchy sequence. – Pedro Nov 06 '17 at 04:04
  • $1/n$ is a sequence converging to $0$. So, it is a Cauchy sequence converging to $0$. But $0$ is not a point of $(0,1)$, so the fact that $f(a_n)$ doesn't converge to $f(0)$ does not tell us anything. – user5826 Nov 06 '17 at 04:09
  • @AlJebr Explicitly and rigorously, what do you mean by "A function is continuous iff it preserves convergent sequences"? – Pedro Nov 06 '17 at 04:15
  • Let $f$ be a function defined on $D$. Then $f$ is continuous if and only if for every sequence $x_n \to x_0 \in D$, we have $f(x_n) \to f(x_0)$. – user5826 Nov 06 '17 at 04:19
  • For $f(x)=1/x$ on $(0,1)$, we know $f$ is not continuous at $0$ because $0$ is not even in the domain. – user5826 Nov 06 '17 at 04:20
  • @AlJebr see my edit. Step 2 is wrong, right? In the sentences that you wrote, the fact that the Cauchy sequence converges to a point of the domain is a hidden assumption (not fulfilled in general, as shown by that example). – Pedro Nov 06 '17 at 04:35
  • Ok, so the problem arises at limit points? For the $f(x) =1/x$ defined on $(0,1)$ example, although $0$ is a limit point and so there exists a sequence $x_n \to 0$, we cannot conclude $\lim f(x_n)$ exists? – user5826 Nov 06 '17 at 04:55
  • @AlJebr Exactly. The argument works if all limit points of the domain belong to the domain (i.e. if the domain is closed). – Pedro Nov 06 '17 at 04:57
  • @reuns exactly. Nice conclusion. – MathMan May 04 '18 at 07:44