Proof of compactness of subset S of a compact set K contained in an open set.

Question

Let $A$ be an open subset $\mathbb{R}^{n}$ and $K\subset A$ is compact. Prove there exist an $r\gt0$ such that the set $$ D=\{y\in \mathbb{R}^{n}: \|y-x\|\leq r \text{ for some }x \in K\}$$ is a compact subset of $A$.

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Let $y\in cl(D)$ (closure of D), there exist a sequence $y_n$ in $D$ which converges to $y$. This implies that $\|y_n-x_n\|\leq r$, now $x_n$ is a sequence in $K$ and must have a convergent subsequence(due to compactness of K) and point of convergence in $K$, say $x$.

We choose the subsequence of $y_n$ in such way to match with numbering of subsequence of $x_n$.

$\|y-x\|\leq \|y_{n_k}-y\|+\|y_{n_k}-x_{n_k}\|+\|x_{n_k}-x\|\leq 2\epsilon+r$

This implies $y\in D$ and $cl(D)=D$. $\blacksquare$

I have doubt whether I'm missing something and I haven't made use of the set $A$ at all and also the existence of $r>0$

Answer 1

You've proved that $D$ is a compact set for any $r$, but you haven't shown that an $r$ exists such that $D\subset A.$

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For each $x\in K$, find $r_x>0$ such that $\displaystyle{N_{r_x}(x)\subseteq A.}$ We can do this since $x\in A,$ and $A$ is open.

Now, let $\displaystyle{U_x=N_{r_x/3}(x).}$

Then $\bigcup_{x\in K} U_x$ is obviously an open cover of $K$, so we must have an $x_1,x_2,\dots,x_n\in K$ such that:

$$K\subseteq \bigcup_{i=1}^{n} U_{x_i}\subseteq A$$

Now, let $\displaystyle{r=\frac 13\min {r_{x_i}}.}$ This is positive since it is the minimum of a finite set of positive reals.

We will show that this $r$ works.

If $y\in D$ then there is some $y_0\in K$ such that $|y-y_0|\leq r.$

Since the $U_{x_i}$ cover $K,$ there must be an $x_i$ such that $y\in U_{x_i},$ which means $\displaystyle |y_0-x_i|<r_{x_i}/3.$

Now $r<\frac{r_{x_i}}{3}$ so we have that:

$$|y-x_i|\leq |y-y_0|+|y_0-x_i|<\frac{2}{3}r_{x_i}$$

Which means that $$y\in N_{r_{x_i}}(x_i)\subseteq A.$$

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You definitely need compactness.

If $A=\{(x,y)\mid |xy|<1\}$ and $K=\{(0,y)\mid y\in\mathbb R\},$ then $A$ is open and $K$ is closed, but you don't have a value $r$.

If $A=(-1,1)\subseteq \mathbb R^1$ and $K=[0,1)$, then $K$ is bounded, and even closed as a subset of $A$, but is not closed in $\mathbb R$, hence not compact, and again there is no $r.$

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Another approach, using Cantor's intersection theorem.

Let $D_r$ be the set you get for a value $r>0.$ You've already shown that $D_r$ is compact for all $r.$

Now, assume that there is no $D_r\subseteq A.$ Then let $$C_r=D_r\setminus A=D_r\cap(\mathbb R^n\setminus A)$$

Since $\mathbb R^n\setminus A$ is closed, $C_r$ is a closed subset of $D_r,$ hence compact. We also have $C_{r}\subseteq C_{r'}$ if $r<r'.$ And, by our assumption, the $C_r$ are non-empty.

So we have that $\bigcap_{k=1}^{\infty} C_{1/k}$ is the intersection of a nested chain of non-empty compact sets, and hence is non-empty by Cantor's intersection theorem. So there is a $y$ such that $y\in C_{1/k}$ for all $k.$

Since $C_{1/k}\subseteq D_{1/k},$ this means that for each $k>0$ there is an $x_k\in K$ such that $|y-x_k|\leq\frac{1}{k}.$ But then $x_k\to y,$ and thus $y\in K$, contradicting that $y\not\in A.$

This means some $C_{1/k}$ is empty, which means that $D_{1/k}\subseteq A.$

Answer 2

From an advanced standpoint we can use some basic results about compact spaces and metric spaces:

(1). If $K$ is a non-empty compact subset of a space $X$ and if $f:X\to \Bbb R$ is continuous then $\min \{f(x):x\in K\}$ exists.

(2). If $(X,d)$ is a metric space and $V$ is a non-empty subset of $X$ then the function $f(x)=\inf \{d(x,y):y\in V\}$ is continuous from $X$ to $\Bbb R.$

(3). If $(X,d)$ is a metric space and $K$ is a closed bounded subset of $X$ and $r\in \Bbb R^+$ then $D= \{z\in X: \exists x\in K\;(d(z,x)\leq r\}$ is closed and bounded.

For the Q, supposing $K\ne \emptyset$ and $A\ne \Bbb R^n$:

Let $f(x)=\inf \{\|x-z\|: z\in \Bbb R^n\setminus A\}$ for $x\in \Bbb R^n.$ Since $A$ is open, $f(x)\ne 0$ for all $ x\in A,$ so $f(x)\ne 0$ for all $x\in K.$

By (1), there exists $x_0\in K$ with $f(x_0)=r=\min \{f(x):x\in K\}.$ And $r\ne 0$ because $x_0\in A.$ So $\forall x\in K \;(f(x)\geq r).$

Let $D= \{z\in \Bbb R^n: \exists x\in K\;(\|z-x\|\leq r/2\}.$ Now $K$ is closed and bounded, so by (3), $D$ is closed and bounded, and in $\Bbb R^n$ this implies compactness.

If $z\in \Bbb R^n \setminus A $ then for any $x\in K$ we have $\|z-x\|\geq f(x)\geq f(x_0)=r ,$ so $z\not \in D$. Therefore $D\subset A.$

For the trivial cases: If $K$ is empty or if $A=\Bbb R^n$ let $D=K.$

Remarks: (1) is a corollary of the result that a continuous image of a compact space is compact, so $\{f(x):x\in K\}$ is a non-empty compact sub-space of $\Bbb R,$ so it must have a minimum..... In (3), let $g(z)=\inf \{d(z,x):x\in K\}.$ Since $K$ is closed , we have $X \setminus D=g^{-1}(r,\infty),$ which is open by (2) so $D$ is closed.