analysis - Proof on showing a uniformly continuous function has limit at every cluster point of the domain

The question is as follows:

Given:

(a) f is uniformly continuous on a subset D of $\mathbb R^n$

and (b) $x_0$ is a cluster point of D

Show: The limit of f(x), as x approaches $x_0$, exists in D

Here are my attempts:

Attempt 1: Direct proof

1/ By definition of uniform continuity, f is uniformly continuous on D

<==> for any $\epsilon$ > 0, there exists $\delta$ such that for any $x_n$ and $y_n$ in D,

we have d($x_n$,$y_n$) < $\delta$ ==> d(f($x_n$), f($y_n$)) < $\epsilon$

2/ By definition of a cluster point, $x_0$ is a cluster point of D ==> let N be a neighborhood about $x_0$, this neighborhood intersects D by at least one point k in D and k is different from $x_0$

Then.... I don't know how to proceed >_<
I had a feeling that I have to use uniform continuity (thus continuity) to say for any x in the intersection of the neighborhood N about $x_0$ and D, f(x) should be in the neighborhood M about $f(x_0)$ But... I'm not sure.

Attempt 2: Proof by Contradiction

1. Assume by contradiction that such limit doesn't exist.




2. Then by negating the definition of limit, I claim there is a sequence {$x_n$} such that {$x_n$} approaches $x_0$, but {f($x_n$)} doesn't approach to any limit value L in D




3. Since {$x_n$} approaches $x_0$, the sequence is Cauchy
   Thus sequence {f($x_n$)} is also Cauchy




4. Then since Cauchy sequences are convergent, the sequence {f($x_n$)} converges to some N in D




5. But this is not true since I claim no such limit N exists.
   Thus the original conclusion should be true.

Would someone please help me on this problem?

Thank you very much ^_^

Answer

What is your codomain?
Let’s say it’s $R^n$ and let $f \colon D → R^n$ be uniformly continuous.

You have to show that there is a limit $L ∈ R^n$ such that $f(x) \overset{x→x_0}\longrightarrow L$, that is:

For every sequence $(x_n)_{n ∈ ℕ}$ in $D$ converging to $x_0$ you have $f(x_n) \overset{n → ∞}\longrightarrow L$.

Note that this limit has to be the same for every such sequence.
So you still have to show uniqueness.

Alternatively you can show:

(1) For every $ε > 0$ there is a $δ > 0$ such that $f(U_δ(x_0)) ⊂ U_ε(L)$.

Do this by first showing that:

(2) For every $ε > 0$ and $δ > 0$ there is a $x_{ε,δ} ∈ U_δ(x_0)$ such that $f(x_{ε,δ}) ∈ U_ε (L)$.

This you have already done.
It is the choice of your $L$:

The crucial point here is that, since $x_0$ is a limit point, you have a sequence in $D$ converging to $x_0$ whose image sequence still converges.
You should carry out your argument that the image of that sequence is again Cauchy, using uniform continuity.
Then, by completeness you get your limit $L$.

Next, show (1):
Let $ε > 0$.
Set $η = ε/2$ and choose $δ > 0$ such that for any $x, x' ∈ D$ one has $|f(x') - f(x)| < η$ whenever $|x - x'| < δ$.
Now, for any $x ∈ U_δ(x_0)$ you have $|f(x) - L| ≤ |f(x) - f(x_{η,δ})| + |f(x_{η,δ}) - L| < η + η = ε$ and you’re done.