I'm trying to prove that if $\{x_n\}$ is Cauchy sequence which located in $E$ ($f$ is uniformly continuous) then $\{f(x_n)\}$ is a Cauchy.

let say {$x_n$} is a Cauchy sequence in $E$ and $E \to R$ and $f$ is uniformly continuous

since {$x_n$} is Cauchy sequence so there should be exist

$|p_n - p_m| < \epsilon$

and also its uniformly continuous on $E$, this imply that

$|f(p_n)- f(p_m)| $ also has the $\delta$ such as $|f(pn)- f(pm)| < \delta$

then this can attribute to say that

{${f(x_n)}$} is the Cauchy sequence

please take a look my proof and correct it

  • $\begingroup$ What do you mean by $E\rightarrow R$ is continuous? Is this a function? I guess you mean $f$ is uniformly-continuous? $\endgroup$ – user99680 Nov 13 '13 at 18:53
  • You have the right idea, but you need to be precise with using definitions and demonstrating the required properties.

  • You also need to be consistent with your notation: why do you switch from $X_n$ to $p_n$?

  • You should also avoid using the word "its" unless it's very clear what you are referring to.

  • State what you will prove at the beginning of your proof.


Choose any $\epsilon > 0$. We will show that there exists $N>0$ such that if $m,n>N$ then $|f(x_m) - f(x_n)| < \epsilon$.

Given any $\delta > 0$, Since $\{x_n\}$ is a Cauchy sequence, there exists $N$ so that $n,m>N$ implies $$ |x_n - x_m| < \delta. $$

Since f is uniformly continuous on $E$, given any $\epsilon > 0$ there exists $\delta > 0$ so that $$ |x-y| < \delta $$ implies $$ |f(x) - f(y)| < \epsilon. $$

So, if $m,n > N$, then we have $|x_m - x_n| < \delta$ which implies $|f(x_m) - f(x_n)| < \epsilon$.


If I understood well, if f is uniformly-continuous, then for any $n$, here is a $\delta(n) $ so that :

$|f(x_n)-f(x_m)|< 1/n$ when $|x_n-x_m|<\delta$.

Since {$x_n$} is Cauchy, $\delta$ can be made as small as possible, for specifically, for any $n_0$, there is an $m$ with $|x_k-x_j|<\delta (n_0)$ implies, by uniform continuity of $f$, that: $$|f(x_k)-f(x_j)|<1/n_0$$

Note that uniform continuity is necessary: the function $f: \mathbb Q \rightarrow \mathbb Q $ defined by $f(x)=\frac {1}{x-\pi}$, and consider the sequence $a_n$ where $a_n$ is the first $n$ terms of the decimal expansion of $\pi$. Now, $a_n \rightarrow \pi$ , but $f(a_n)\rightarrow \infty $.


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