Proposition: Let $X$ be a subset of $R$, let $f:X\to R$ be a uniformly continuous function, and let $x_0$ be an adherent point of $X$. Then $\lim_{x\to x_0 ;x\in X} f ( x)$ exists.

Proof Take any sequence $(a_n)_{n=1}^\infty$ where $a_n\in X$ and it converges to $x_0$. We know such a sequence exists because $x_0$ is an adherent point of $X$.

Since $(a_n)$ is a convergent sequence, then its cauchy. $f$ is a uniform continuous function,hence, if $(a_n)$ is cauchy, $(f(a_n))$ is also cauchy. Again, because $(f(a_n))$, then its convergent and hence the limit exists.

Is my proof correct?

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    $\begingroup$ You have to prove that this limit is independent of the approximating sequence $(a_n)$. $\endgroup$ – Jose27 Jan 24 '15 at 6:21
  • $\begingroup$ @Jose27 I took any convergent sequence, is that enough? $\endgroup$ – MAS Jan 24 '15 at 6:30

You have just proved limit exists for a specific convergent sequence $x_n\to x_0$, you have to show then the limit is independent of the convergent sequence. That is if $x_n\to x_0,y_n\to x_0$, then $\lim f(x_n)=\lim f(y_n)$, which should be easy if you make use of the uniform continuity of $f$ on $X$ and pass to limit.

Respond to J.G's comment:

Given $x_n\to x_0,y_n\to x_0$, we know $\lim f(x_n),\lim f(y_n)$ exists and aim to show $\lim f(x_n)=\lim f(y_n)$. Consider sequence $z_n$ defined as $(x_1,y_1,x_2,y_2,\cdots)$ which is a combination of $(x_n),(y_n)$. Then we know $z_n\to x_0$ and by similar argument, we know $\lim f(z_n)=l\in \mathbb{R}$. Since $f(x_n),f(y_n)$ are two subsequence of a convergence sequence $\lim f(z_n)$, we know they must convergent to the same limit, hence $\lim f(x_n)=\lim f(y_n)$.

  • $\begingroup$ I should use the fact if (a_n) and (b_n) are equivalent, then their image under f are equivalent? $\endgroup$ – MAS Jan 24 '15 at 7:14
  • $\begingroup$ @MAS Yes, I think that is the idea. $\endgroup$ – John Jan 24 '15 at 7:17
  • $\begingroup$ @JohnZHANG I had a similar argument as MAS, but I can't show the part you explain. To know that we have not yet seen that two sequences are equivalent. Are you able to complete the proof in your comment? $\endgroup$ – user230283 Feb 1 '16 at 22:24
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    $\begingroup$ @J.G I have edited the answer. $\endgroup$ – John Feb 2 '16 at 3:47

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