3.11. Special Topics

This section covers details on specific functions from the perspective of metric spaces

3.11.1. Distance from a Point

Theorem 3.122 (Distance function is uniformly continuous)

Let $(X,d)$ be a metric space. Fix some point $a\in X$. Define a function $f:X\to \mathbb{R}$ as:

$$f(x)=d(x,a)\mathrm{\forall }x\in X.$$

Then, $f$ is uniformly continuous on $X$.

Proof. Let $x,y\in X$. Recall from triangle inequality that:

$$|d(x,a)-d(y,a)|\le d(x,y).$$

1. Let $ϵ>0$.

2. Choose $\delta =ϵ$.

3. Assume $d(x,y)<\delta$.

4. Then

   $$|f(x)-f(y)|=|d(x,a)-d(y,a)|\le d(x,y)<\delta =ϵ.$$

5. Thus, $d(x,y)<\delta ⟹|f(x)-f(y)|<ϵ$.

6. Thus, $f$ is uniformly continuous.

3.11.2. Distance from a Set

Recall from Definition 3.5 that the distance between a point $x\in X$ and a set $A\subseteq X$ is given by:

$$d(x,A)=inf\{d(x,a)\mathrm{\forall }a\in A\}.$$

Proposition 3.32 (Distance from a singleton set)

For every $x,y\in X$,

$$d(x,\{y\})=d(x,y).$$

Proof. Let $A=\{y\}$ be a singleton set. Then

$$d(x,A)=d(x,\{y\})=inf\{d(x,y)\}=d(x,y).$$

Proposition 3.33 (Distance from a subset)

Let $\varnothing \ne A\subseteq B\subseteq X$. Then, for any $x\in X$

$$d(x,A)\ge d(x,B).$$

Proof. We note that since $A\subseteq B$, hence:

$$\{d(x,a)\mathrm{\forall }a\in A\}\subseteq \{d(x,a)\mathrm{\forall }a\in B\}.$$

The infimum over a larger set is smaller. Hence,

$$d(x,B)\le d(x,A).$$

Proposition 3.34 (Distance from a containing subset)

Let $x\in A\subseteq X$. Then

$$d(x,A)=0.$$

Proof. We note that:

$$d(x,x)\in \{d(x,a)\mathrm{\forall }a\in A\}$$

since $x\in A$.

Thus,

$$d(x,A)=d(x,x)=0.$$

Theorem 3.123 (Continuity of set distance function)

Let $A\subseteq X$ be nonempty. Then, the function $f:X\to \mathbb{R}$ given by:

$$f(x)=d(x,A)=inf\{d(x,a)\mathrm{\forall }a\in A\}$$

is continuous.

We provided a proof in Example 3.16. We rephrase it here.

Proof. Let $a\in X$. We shall show that $f$ is continuous at $a$.

1. Let $r>0$.

2. Consider the open ball/interval $V=B(f(a),r)$ in $\mathbb{R}$.

3. Consider the open ball in $X$ given by $U=B(a,r)$.

4. Let $x\in U$ and let $y\in A$.

5. Then, using triangle inequality:

   $$d(a,y)+d(x,a)\ge d(x,y)$$

   and

   $$d(a,x)+d(x,y)\ge d(a,y).$$

6. Taking the infimum over $y\in A$ in both the inequalities, we get:

   $$d(a,A)+d(x,a)\ge d(x,A)$$

   and

   $$d(x,a)+d(x,A)\ge d(a,A).$$

7. Rewriting it further, we get:

   $$d(x,A)\le d(a,A)+d(x,a)$$

   and

   $$d(a,A)-d(x,a)\le d(x,A).$$

8. Since $d(a,x)<r$, hence, we obtain:

   $$d(a,A)-r<d(x,A)<d(a,A)+r.$$

9. Substituting $f(x)=d(x,A)$ and $f(a)=d(a,A)$,

   $$f(a)-r<f(x)<f(a)+r.$$

10. Thus, $f(x)\in B(f(a),r)=V$.

11. In other words, $f(U)\subseteq V$.

We have shown that for every $r>0$, there exists $\delta =r$ such that $x\in B(a,\delta )$ implies $f(x)\in B(f(a),r)$. Thus, $f$ is continuous at $a$.

Theorem 3.124 (Open neighborhoods of a set)

Let $A\subseteq X$ be nonempty. Let $ϵ>0$ and consider the set $A_{ϵ}$ given by

$$A_{ϵ}=\{x\in X|d(x,A)<ϵ\}.$$

Then $A_{ϵ}$ is open for every $ϵ>0$.

Proof. We established in Theorem 3.123 that $f(x)=d(x,A)$ is continuous. The set $A_{ϵ}$ is the inverse image of the open interval $(-ϵ,ϵ)$ in $\mathbb{R}$. Since $f$ is continuous, hence the inverse image of an open interval is an open set.

Following is a direct proof.

1. Let $x\in A_{ϵ}$.

2. Then, there is $y\in A$ such that $d(x,y)<ϵ$.

3. Let $\delta >0$ be small enough such that $d(x,y)+\delta <ϵ$.

4. Consider the open ball $B(x,\delta )$.

5. For all $z\in B(x,\delta )$, we have:

   $$d(z,y)\le d(z,x)+d(x,y)<\delta +d(x,y)<ϵ.$$

6. Thus, $d(z,A)<ϵ$ since $y\in A$.

7. Thus, $z\in A_{ϵ}$.

8. Thus, $B(x,\delta )\subseteq A_{ϵ}$.

9. Thus, $A_{ϵ}$ is open.

Theorem 3.125 (Closure and set distance)

Let $A\subseteq X$ be a nonempty set. Then, for any $x\in X$, $d(x,A)=0$ if and only if $x\in \mathrm{cl}A$.

In other words, the distance of a point from a set is zero if and only if the point is a closure point of the set.

Proof. Let $x\in \mathrm{cl}A$.

1. Then either $x\in A$ or $x\in \mathrm{bd}A$.

2. If $x\in A$, then by Proposition 3.34, $d(x,A)=0$.

3. Now, assume $x\notin A$ and $x\in \mathrm{bd}A$.

4. Let $r>0$.

5. Since $x\in \mathrm{bd}A$, hence there exists $y\in A$ such that $d(x,y)<r$.

6. Therefore,

   $$d(x,A)=inf\{d(x,y)|y\in A\}<r.$$

7. Since this is true for every $r>0$, hence $d(x,A)=0$.

Now assume that $x\notin \mathrm{cl}A$.

1. Thus, there exists an open ball $B(x,r)$ for some $r>0$ such that $B(x,r)\cap A=\varnothing$.

2. Therefore, for every $a\in A$, $d(x,a)\ge r>0$.

3. Therefore,

   $$d(x,A)=inf\{d(x,a)|a\in A\}\ge r>0.$$

Corollary 3.4

Let $A\subseteq X$ be a nonempty closed set. Then, for any $x\in X$, $d(x,A)=0$ if and only if $x\in A$.

Proof. A closed set contains all its closure points. Thus, $d(x,A)=0$ if and only if $x$ is a closure point of $A$ if and only if $x\in A$.

Theorem 3.126 (Distance minimizer in a compact set)

Let $A$ be a nonempty compact subset of $X$. Let $x\in X$. Then, there exists an $a\in A$ such that

$$d(x,A)=d(x,a).$$

Proof. For a fixed $x\in X$, consider the real valued function $f:X\to \mathbb{R}$ given by:

$$f(y)=d(y,x)\mathrm{\forall }y\in X.$$

Then, $f$ is continuous as it is a distance function from a singleton set $\{x\}$.

Hence, $f(A)$ attains a minimum value at some $a\in A$. See Theorem 3.85.

Theorem 3.127 (Distance minimizer in a closed set)

Let $X$ be a Euclidean metric space. Let $A\subseteq X$ be a nonempty closed set. Let $x\in X$. Then, there exists an $a\in A$ such that

$$d(x,A)=d(x,a).$$

In other words, the infimum of the distance between $x$ and points of $A$ is realized at a point in $A$.

Proof. Since $A$ is nonempty, we can pick some $b\in A$ and compute $r=d(x,b)$. Clearly,

$$d(x,A)\le d(x,b)=r.$$

Consider the set

$$C=\{a\in A|d(x,a)\le d(x,b)=r\}=A\cap B[x,r].$$

In other words, $C$ is the intersection of $A$ and the closed ball $B[x,r]$ centered at $x$ and of radius $r$. Since $A$ is closed and $B[x,r]$ is closed, hence $C$ is closed.

It is easy to see that

$$d(x,A)=d(x,C).$$

For any points $u,v\in C$, we have:

$$d(u,v)\le d(u,x)+d(x,v)\le r+r=2r.$$

Thus, $C$ is closed and bounded. By Heine-Borel theorem, $C$ is compact.

Now, for a fixed $x\in X$, consider the function $f:X\to \mathbb{R}$ given by:

$$f(y)=d(y,x)\mathrm{\forall }y\in X.$$

Then, $f$ is continuous.

1. By Theorem 3.78, $f(C)$ is compact (continuous images of compact sets are compact).

2. But $f(C)\subseteq \mathbb{R}$.

3. Hence, $f(C)$ is closed and bounded as $\mathbb{R}$ is Euclidean (Heine-Borel theorem).

4. Hence, $f(C)$ attains a minimum value at some $a\in C$. See Theorem 3.85.

3.11.3. Distance Between Sets

Recall from Definition 3.6 that the distance between two subsets of a metric space is given by

$$d(A,B)=inf\{d(a,b)|a\in A,b\in B\}.$$

Theorem 3.128 (Distance between disjoint compact and closed sets)

Let $(X,d)$ be a metric space. Let $A,B\subseteq X$ be disjoint (i.e., $A\cap B=\varnothing$). Assume that $A$ is compact and $B$ is closed.

Then, there is $\delta >0$ such that

$$|x-y|\ge \delta \mathrm{\forall }x\in A,y\in B.$$

In other words, the distance between $A$ and $B$ is nonzero; i.e., $d(A,B)>0$.

Proof. We prove this by contradiction.

1. Assume that there is no such $\delta >0$.

2. Then there exist sequences $\{x_n\}$ of $A$ and $\{y_n\}$ of $B$ such that

   $$\underset{n\to \mathrm{\infty }}{lim}|x_n-y_n|=0.$$

3. Since $A$ is compact, hence $\{x_n\}$ has a convergent subsequence, say $\{x_{n_k}\}$ which converges to some $x\in A$.

4. Now,

   $$|x-y_{n_k}|\le |x-x_{n_k}|+|x_{n_k}-y_{n_k}|.$$

5. Since $\{x_n-y_n\}$ is a convergent sequence, hence $\{x_{n_k}-y_{n_k}\}$ also converges to $0$.

6. Thus,

   $$\underset{n\to \mathrm{\infty }}{lim}|x-y_{n_k}|\le \underset{n\to \mathrm{\infty }}{lim}|x-x_{n_k}|+\underset{n\to \mathrm{\infty }}{lim}|x_{n_k}-y_{n_k}|=0+0=0.$$

7. Thus, $\{y_{n_k}\}$ is a convergent sequence of $B$ converging to $x$.

8. Since $B$ is closed, hence $x\in B$.

9. But then, $x\in A\cap B$.

10. Thus, $A\cap B\ne \varnothing$ which is a contradiction.