Let $(S,d)$ and $(S^*, d^*)$ be metric spaces. If $f:S \to S^*$ is uniformly continuous and if $(s_n)$ is a Cauchy sequence in $S$, then $(f(s_n))$ is a Cauchy sequence in $S^*$.

$f:S \to S^*$ is uniformly continuous: $(\forall \varepsilon >0)( \exists > 0)(\forall s,t \in S)$ $d(s,t) < \delta \implies d(f(s), f(t)) < \varepsilon$.

$(s_n)$ Cauchy: $(\forall \varepsilon > 0)(\exists N \in \mathbb{N})$ such that $\forall n,m \geq N$ we have $d(s_n,s_m)) < \varepsilon$.

Now I need to show that $(\forall \varepsilon > 0)(\exists N \in \mathbb{N})$ such that $\forall n,m \geq N$ we have $d(f(s_n),f(s_m)) < \varepsilon$.

Since $f$ is uniformly continuous, $\delta$ will depend on $\varepsilon$ only. Then I separate the problem into two cases:

Suppose $\varepsilon < \delta$: Since I know $(s_n)$ is Cauchy, I can take the same $N$ to get $d(s_n,s_m) < \varepsilon < \delta$ which implies $d(f(s_n),f(s_m)) < \varepsilon$ by uniform continuity.

Suppose $\delta \leq \varepsilon$: Since $f$ is uniformly continuous, $d(s,t) < \delta \implies d(f(s),f(t)) < \varepsilon$ will hold for all $s,t \in S$. So let $N=1$, then $d(s_n,s_m) < \delta$ which implies $d(f(s_n),f(s_m)) < \varepsilon$.

Could someone give me feedback on my proof?

Edit For every $\varepsilon > 0$, since $f$ is uniformly continuous, there exists a $\delta >0$ such that for all $s, t \in S$, $d(s, t) < \delta \implies d(f(s),f(t)) < \varepsilon$. Then for any $\delta >0$ there exists an $N \in \mathbb{N}$ such that for all $n,m \geq N$, $d(s_n, s_m) < \delta$. By uniform continuity, $d(s_n, s_m) < \delta \implies d(f(s_n),f(s_m)) < \varepsilon$.


1 Answer 1


The first part (case $\varepsilon<\delta$) is correct, but I can't follow the second part. The conclusion $N=1$ is clearly wrong for arbitrarily small $\varepsilon$'s.

In the definition of uniform continuity, the presence of $\delta$ is always around $S$ and $\varepsilon$ is around $S^*$. So, we want to prove $f(s_n)$ is Cauchy:

Let's assume an $\varepsilon>0$ is given. For this we can choose a $\delta$, and for this $\varepsilon':=\delta$ we can choose an $N$ for $s_n$ by the Cauchy property.

  • $\begingroup$ Why are we allowed to choose $\delta$ and $N$? In the definitions I only know that there exists $\delta$ and $N$. $\endgroup$
    – Student
    Jun 26, 2013 at 23:12
  • $\begingroup$ How about this: for every $\varepsilon >0$, there exists a $\delta$ so that $f$ is uniformly continuous. For any $\delta >0$ since $(s_n)$ is Cauchy, there exists an $N$ such that $d(s_n,s_m) < \delta$. So by uniform continuity, $d(f(s_n),f(s_m)) < \varepsilon$. $\endgroup$
    – Student
    Jun 26, 2013 at 23:21
  • 1
    $\begingroup$ No, First uniform continuity gives us a $\delta$. Then we feed the definition of Cauchy sequence by this $\delta$ as $\varepsilon$. $\endgroup$
    – Berci
    Jun 26, 2013 at 23:30
  • $\begingroup$ Isn't that what I have written above? $\endgroup$
    – Student
    Jun 26, 2013 at 23:31
  • $\begingroup$ Sorry I see what you mean now. $\endgroup$
    – Student
    Jun 26, 2013 at 23:36

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