Proof about uniform continuity

Question

I'm stuck on this problem:

Let $C\subset \mathbb{R}^n$ be closed and unbounded.

Suppose $f:C\to\mathbb{R}^m$ is continuous and such that $\lim_{x\to\infty} f(x)$ exists and is finite.

Show that $f$ is uniformly continuous on $C$.

I tried to prove it using the definition of limit and continuity but it didn't get me anywhere. I appreciate any advice.

Answer 1

Let $\epsilon > 0$ and $L$ be the limit of $f$ at $\infty$. Choose $M$ so that $|x| > M \Rightarrow |f(x) - L| <\epsilon$. Notice that $\{x\in C| |x|\le M\}$ is compact so $f$ is uniformly continous there. Can you do the rest?

Answer 2

Suppose $f$ is continuous but not uniformly continuous on $[a,b]$. Then, there exists an $\varepsilon_0 > 0$ such that $$|x - y| < \delta \ \Rightarrow \ |f(x) - f(y)| < \varepsilon_0 \ \ \text{is false} \ \forall \delta$$ For $\delta = \frac{1}{N}$, we can find two points $x_n,y_n\in [a,b]$ such that $$|x_n - y_n| < \frac{1}{N} \ \ \text{and} \ \ |f(x_n) - f(y_n)| \geq \varepsilon_0$$ This gives us two sequences $\{x_n\}$ and $\{y_n\}$ both lying in $[a,b]$, and with $|x_n - y_n| \rightarrow 0$. By the Bolzano-weierstrass theorem, there exists a subsequence $\{x_{n_{k}}\}$ of $x_n$, and a point $x\in [a,b]$, such that $x_{n_{k}}\rightarrow x$ (this is where we are using the fact the interval is closed). Now, put $\tilde{x}_{n_{k}} = x_{n_{k}}$, and $\tilde{y}_{n_{k}} = y_{n_{k}}$. So, $\tilde{x}_{n_{k}}\rightarrow x$, and $\tilde{y}_{n_{k}}$ is a subsequence of the original sequence $y_n$.

By the same reasoning, there exisits $y\in [a,b]$ and a subsequence $\{Y_{k_{s}}\}$ with $\{Y_{k_{s}}\}\rightarrow y$. Put $\hat{x_{s}} = \tilde{x}_{k_{s}} $ , $\hat{y_{s}} = \tilde{y}_{k_{s}} $. Then $\hat{x_s}$ is a subsequence of a convergent subsequence and hence also converges to $x$. Since $\hat{y}_s\rightarrow y$, and we have $$|\hat{x}_s - \hat{y}_s| \rightarrow 0$$ we must have $$|f(\hat{x}_s) - f(\hat{y}_s)|\rightarrow 0$$ This contradicts the assumption that is bounded away from zero. Therefore, our assumption that $f$ is not uniformly continuous is incorrect.