Induced Metric Space

Definition

Let $(X,d)$ be a Metric Space and let $\varnothing \ne Y\subset X$. Then $(Y,d{|}_Y)$ is a metric space. Indeed the axioms for a metric space hold in the subset.

Definition (Open/Closed Relative)

A subset $A\subset Y$ is said to be open relative to $Y$ if $A$ is open in the metric space $(Y,d)$. Similarly, $A\subset Y$ is said to be closed relative to $Y$ if $A$ is closed in the Metric Space.

Example 1:

Open rel $Y$ could be different than open in $X$.

Consider $\mathbb{R}$ with the standard metric, $Y=(0,1]\ne \varnothing$, $A=(1/2,1]$. $A$ is not open in $X$. However, $A$ is open relative to $(0,1]$. Indeed, if $x\in (1/2,1)$, then $\exists r>0$ such that

$$(x-r,x+r)\subset (1/2,1)\subset (1/2,1]=Y$$

The remaining case is when $x=1$. Then $(1/2,1]=N_{1/2}(1)$ in $Y$ since the neighborhood only samples points in $Y$. So, $\{y\in Y\mid |y-1|<1/2\}=N_{1/2}(1)$ and it is open.

Remark (Relative Neighborhood)

Indeed,

$$N_{1/2}(1)=(1/2,1]=\{y\in Y\mid |y-1|<1/2\}=\{x\in X\mid |y-1|<1/2\}\cap X$$

Remark (More Generally)

Let $(X,d)$ be a Metric Space. Then $\varnothing \ne Y\subset X$ and $p\in Y$ and $r>0$. So,

$$\begin{array}{rl}{N}_r^Y(p)& :=\{y\in Y\mid d(y,p)<r\}\\ & =\{x\in X\mid d(x,p)<r\}\cap Y\\ & =N_r^X(p)\cap Y\end{array}$$

Lemma (Open/Closed by Above)

Let $\varnothing \ne Y\subset X$ and $A\subset Y$.

1. $A$ is open rel $Y$ $⟺\exists O\subset X$ that is open such that $A=O\cap Y$.
2. $A$ is closed rel $Y⟺\exists F\subset X$ closed such that $A=F\cap Y$.

Proof (1):

Find $O$ that is open in $X$ such that $A=O\cap Y$.

• If $A=\varnothing$, then we are done.
• If $A\ne \varnothing$, then $\forall p\in A,\exists r_p>0$ such that

$$p\in N_{r_p}^Y=\{y\in Y\mid d(p,y)<r_p\}\subset A$$

Recall that that this is equal to $N_{r_p}^X(p)\cap Y$ from the remark. Now, let

$$O=\bigcup _{p\in A}{N}_{r_p}^X(p)$$

Note $O$ is a union of open sets in $X$ so it is open in $X$ by this lemma. Also note that $O\cap Y\subset A$ . Indeed,

$$\begin{array}{rlr}O\cap Y& =\left(\bigcup _{p\in A}{N}_{r_p}^X(p)\right)\cap Y& \\ & =\bigcup _{p\in A}\left(N_{r_p}^X(p)\cap Y\right)& \\ & =\bigcup _{p\in A}{N}_{r_p}^Y(p)& \subset A\end{array}$$

The second inequality comes from DeMorgan’s Law.

For the reverse direction, we have that $A\subset O\cap Y$ because $\forall p\in A\subset Y,p\in N_{r_p}^X(p)$ such that $p\in O$. Thus, $O$ is open in $X$ and $O\cap Y=A$

Proof (2):

TODO

Lemma (Bounded)

Let $A\subset \mathbb{R}$ be nonempty and bounded. Then by the Axiom of Completeness we have that $\sup (A)\in \overline{A}$ and $\inf (A)\in \overline{A}$. Supremum and Infimum

Proof: If $\sup (A)\in A$, then $\sup A\in \overline{A}$ as $\overline{A}=A\cup A^{\prime }$. Assume $\sup (A)\notin A$. Let $\ell :=\sup (A),r>0$.

Goal: WTS $N_r(\ell )\backslash \{\ell \}\cap A\ne \varnothing$. But this is equivalent to $((\ell -r,\ell +r)\backslash \{\ell \})\cap A=\varnothing$. Now, $\ell =\sup (A)$ where $\ell -r<\ell$. So $\ell -r$ cannot be an upper bound for $A$. So, $\exists a\in A$ such that $\ell -r<a<\ell$ because $\ell \notin A,a\in A$ so $a<\ell$.

 (-------|--o-----)
l-r      a  l    l+r

$⟹a\in (\ell -r,\ell +r)$ and as $a\in A$, really, $a\in (\ell -r,\ell +r)\cap A$ and $a\ne \ell$. $⟹\ell \in A^{\prime }⟹\ell \in \overline{A}$

If $\inf (A)\in A$ then $\inf (A)\in \overline{A}$. Suppose otherwise, $\inf (A)\notin A$. Let $\ell =\inf (A)$ for $r>0$. Then

$$\ell <\ell +r$$

so $\ell +r$ cannot be a lower bound for $A$. Indeed, $\exists a\in A$ where $\ell <a<\ell +r$. $⟹a\in (\ell -r,\ell +r)\cap A,a\ne \ell$ $⟹\ell \in A^{\prime }⟹\ell \in \overline{A}$