# If continuity preserves convergence, and Cauchy sequences are convergent sequences, why do we need uniform continuity to preserve Cauchy sequences?

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?

• 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 6 '17 at 3:10
• Think of $f(x) = 1/x$ on $(0,1)$ and $a_n = 1/n$. – amsmath Nov 6 '17 at 3:18
• 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? – Al Jebr Nov 6 '17 at 3:38
• "All Cauchy sequences are convergent" False. – zhw. Nov 6 '17 at 3:42
• Ok, I'm considering $\mathbb R$ only. – Al Jebr Nov 6 '17 at 3:44

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

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 $L$.

• (we need continuity on the closed set $\overline{D(f)}$, for $D(f)$ bounded it is the same as uniform continuity) – reuns Nov 6 '17 at 3:34
• @reuns Thanks, I edited my post. – Pedro Nov 6 '17 at 3:40
• But what is wrong with the implications: In $\mathbb R$, $x_n$ Cauchy implies $x_n$ convergent implies $f(x_n)$ convergent implies $f(x_n) Cauchy? Doesn't this mean continuous functions preserve Cauchy? – Al Jebr Nov 6 '17 at 3:48 • @AlJebr Nothing. See the Additional comment here. – Pedro Nov 6 '17 at 3:50 • @AlJebr I must say "for every sequence$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 6 '17 at 3:53