Proof on compact and closed sets

Question

Take $S$ to be compact and $T$ to be a closed set of real numbers. If we assume that $S \cap T = \emptyset$, I want to prove that there is a $\delta > 0$ such that $|s - t| > \delta$ for every $s \in S$ and $t \in T.$ I have a feeling that I will need to use some information on the bounds of $S$ and the bounds of $T$, although I understand that $S$ will be fully bounded while $T$ can only be assumed to be partially bounded (since technically $[0,\infty)$ is closed, for example). That being said, I am not used to setting a lower bound for these problems (tend to bound things above by $\epsilon$), and so I could use some help on how to approach this problem.

Answer 1

Hint:

1. Define for each $t\in T$ a function $d^t:S\to \mathbf R_+$ such that $d^t(s)=|s-t|$. Prove that: (i) $d^t$ is continuous for each $t\in T$ and (ii) $d^t$ achieves a minimum $m(t)>0$.
2. Let $M=\{m(t)\,|\,t\in T\}$. Prove that $\inf M>0$. (further hint: suppose not. Then for each $n\in\mathbf N$ there are $t_n\in T$ and $s_n\in S$ such that $|t_n-s_n|<\frac 1n$. Since the sequence $\{s_n\}\subseteq S$ and $S$ is compact, you can find a convergent subsequence $\{s_{n_k}\}$ of $\{s_n\}$ whose limit is $s\in S$. Show that $\lim_{k\to\infty}t_{n_k}=s$. Then argue that $s\in T$. Conclude.)

Answer 2

We will prove that $X=\{|s-t|:s\in S, t\in T\}$ is closed.

Let $(|s_{n}-t_{n}|)$ be a sequence in $X$ such that $lim|s_{n}-t_{n}|=x$. Then $(s_{n})$ is a sequence in $S$ and $(t_{n})$ is a sequence in $T$. Since $S$ is limited, there is a subsequence $(s_{n_{k}})$ of $(s_{n})$ such that $lim\ s_{n_{k}}=s$; since $S$ is closed, $s\in S$.

Now, note that $$|t_{n_{k}}|=|t_{n_{k}}-s_{n_{k}}+s_{n_{k}}|\leq|t_{n_{k}}-s_{n_{k}}|+|s_{n_{k}}|$$ and, since $|t_{n_{k}}-s_{n_{k}}|$ and $|s_{n_{k}}|$ converges, both are limited. So $|t_{n_{k}}|$ is limited.

Let $(t_{n_{k_{l}}})$ be a subsequence of $|t_{n_{k}}|$ such that $lim\ t_{n_{k_{l}}}=t$. Since $T$ is closed then $t\in T$. We conclude that $$x=lim|s_{n_{k_{l}}}-t_{n_{k_{l}}}|=|s-t|\in X$$ so $X$ is closed.

Take $\delta=inf(X)$. Since $X$ is closed, $\delta\in X$. If $\delta=0$, there is $s\in S,t\in T$ such that $|s-t|=0$, i.e., $s=t$. But $S\cap T=\emptyset$; therefore $|s-t|>\delta$ for all $s\in S, t\in T$.