Basic analysis/topology of a sequence proof

Question

Lemma 2.2

Let $(X,d)$ be a compact metric space and suppose that $U$ is a clopen subset of $X$.

There exists $\delta >0 $ s.t $d(x,y) \geq \delta $ $\forall x \in U $ and $y\in X-U $

Proof attempt

Suppose that it is not true then for all $\delta >0 $ $\exists x \in U $ and $\exists y\in X-U $s.t $d(x,y) < \delta $

Consider $d(x_i,y_i) <\frac{1}{i}$ pick a sub-sequence converging to $x$ then from the sequence $d(x,y_i)< \frac {1}{i} $ pick a sub-sequence converging to $y$. Then this pair $x,y$ has the property that for $\delta >0 $ we have $d(x,y) <\delta$ hence $x=y$ but this is a contradiction as if $ x\in U $ then $x\notin X-U$

Can anyone give a constructive proof? (ie not with negation)

Answer 1

Take $y \in U^c$. For each $n \ge 1$, let $E_n = \{x \in U : d(x,y) \ge \frac{1}{n}\}$. Each $E_n$ is closed and $\cap_{n \ge 1} E_n = \emptyset$. By compactness (and the fact that $E_n \subseteq E_m$ for $n \ge m$), some $E_N \not = \emptyset$.

For $n \ge 1$, let $F_n = \{y \not \in U : \exists x \in U, d(x,y) \le \frac{1}{n}\}$. Each $F_n$ is closed. The above shows that $\cap_n F_n \not = \emptyset$. Therefore, some $F_N \not = \emptyset$. We're done.

Answer 2

Note that $U$ is itself compact (being closed in $X$). Also $\{U\}$ is an open cover of $U$, as $U$ is open. By compactness of $U$ this has a Lebesgue number $\delta>0$. This means that for all $x \in U$, $B(x,\delta)$ is a subset of $U$, and this implies that $d(x, y) \ge \delta$ for all $x \in U, y \in X\setminus U$.

Alternatively, define $\delta$ to be the minimum of the continuous function $f: U \to \mathbb{R}$ given by $f(x) = d(x,X\setminus U)$, which only assumes values $>0$.

Answer 3

Let $F$ be the family of all open balls $B(x,r)$ that are disjoint from $X$ \ $U$, with $x\in U$ and $r>0.$ For every $x\in U$ there exists $B(x,r)\in F$ because $X$ \ $U$ is closed.

Let $C=\{B(x,r/2): x\in U\land B(x,r)\in F\}.$ Then $C$ is an open cover of $U.$ And $U$ is compact. So (assuming $U\ne \phi$) let $D=\{B(x_j,r_j/2):j=1,...,n\}\subset C$ be a cover of $U.$

Let $\delta =\min (r_1/2,..., r_n/2)$.

If $x\in U$ then $x$ belongs to some $B(x_j,r_j/2)\in D$, with $B(x_j,r_j)\cap (X$ \ $U)=\phi,$ so for any $y\in X$ \ $U$ we have $d(y,x)\geq d(y, x_j)-d(x_j,x)\geq r_j-d(x_j,x)>r_j-r_j/2=r_j/2\ge \delta.$