The Relative Metric

Relative Metric

Definition 41.1. Let $(M, d)$ be a metric space and let $X$ be a subset of $M$. The function $d'$ defined by $$d'(x, y) = d(x, y) \quad \text{for } x, y \in X$$ is called the metric for $X$ relative to $M$ or more simply, the relative metric for $X$.

Remark

Remark. Keeping the notation of Definition 41.1, an open ball in $X$ of radius $\varepsilon$ centered at $a$ is the set $$\{ x \in X \, | \, d'(x, a) < \varepsilon \}$$ which we denote $B^X_{\varepsilon}(a)$. Letting $$B_{\varepsilon}(a) = \{ x \in M \, | \, d(x, a) < \varepsilon \}$$ be the open ball in $M$ of radius $\varepsilon$ centered at $a$, we see that $$B^X_{\varepsilon}(a) = B_{\varepsilon}(a) \cap X.$$

Theorem 41.2

Theorem 41.2. Let $M$ be a metric space and let $X$ be a subset of $M$ with the relative metric. Let $Y$ be a subset of $X$.
(i) $Y$ is open in $X$ if and only if $Y = X \cap U$, where $U$ is open in $M$.
(ii) $Y$ is closed in $X$ if and only if $Y = X \cap C$, where $C$ is closed in $M$.

Proof. (i) ($\Longrightarrow$) Suppose that $Y$ is open in $X$. $\forall y \in Y, \exists \varepsilon_y > 0$ such that $B^X_{\varepsilon_y}(y) \subset Y$. Let $U = \bigcup_{y \in Y} B_{\varepsilon_y}(y)$. Then $U$ is open in $M$. 
Thus $$X \cap U = X \cap \left( \bigcup_{y \in Y} B_{\varepsilon_y}(y) \right) \\ = \bigcup_{y \in Y} (X \cap B_{\varepsilon_y}(y)) = \bigcup_{y \in Y} B^X_{\varepsilon_y}(y).$$ Let $y \in Y$. Since $Y$ is open in $X$, $\exists \varepsilon_y > 0$ such that $B^X_{\varepsilon_y}(y) \subset Y$. Then $\bigcup_{y \in Y} B^X_{\varepsilon_y}(y) \subset Y$. 
Note that $y \in B^X_{\varepsilon_y}(y) \subset \bigcup_{y \in Y} B^X_{\varepsilon_y}(y)$. Thus $Y = \bigcup_{y \in Y} B^X_{\varepsilon_y}(y)$, which means that $Y = X \cap U$.
($\Longleftarrow$) Suppose that $Y = X \cap U$, where $U$ is open in $M$. Let $y \in Y$. Then $\exists \varepsilon > 0$ such that $B_{\varepsilon}(y) \subset U$. Note that $y \in X \cap B_{\varepsilon}(y) = B^X_{\varepsilon}(y) \subset X \cap U = Y$. Thus $Y$ is open in $X$.
(ii) ($\Longrightarrow$) Suppose that $Y$ is closed in $X$. Note that $\overline{Y}$ is closed in $M$. Put $C = \overline{Y}$. Clearly, we have $Y \subset X \cap C$. Let $y \in X \cap C$. Then $\exists \{ y_n \}$ such that $y_n \in Y, \forall n$ and $y_n \to y$. Since $Y$ is closed in $X$, $y \in Y$, which means that $Y = X \cap C$. 
($\Longleftarrow$) Suppose that $Y = X \cap C$, where $C$ is closed in $M$. Let $y$ be a limit point of $Y$. Then $\exists \{y_n\}$ such that $y_n \in Y, \forall n$ and $y_n \to y$. Since $y_n \in C, \forall n$, $y$ is a limit point of $C$. Since $C$ is closed, $y \in C$. Since $y_n \in X, \forall n$ and $y_n \to y$, $y \in X$. Thus $y \in X \cap X = Y$, which means that $Y$ is closed in $X$. $\blacksquare$  

Corollary 41.3

Corollary 41.3. (i) Let $X$ be an open subset of a metric space $M$ and let $Y \subset X$. Then $Y$ is open in $X$ $\iff$ if $Y$ is open in $M$.
(ii) Let $X$ be a closed subset of a metric space $M$ and let $Y \subset X$. Then $Y$ is closed in $X$ $\iff$ $Y$ is closed in $M$.

Proof. ($\Longrightarrow$) Suppose that $Y$ is open in $X$. By Theorem 41.2, $Y = X \cap U$ for some open subset $U$ of $M$. Since $X$ and $Y$ are open in $M$, so is $Y$. 
($\Longleftarrow$) Suppose that $Y$ is open in $M$. Let $y \in Y$. Then $\exists \varepsilon > 0$ such that $B_{\varepsilon}(y) \subset Y \subset X$. Note that $B^X_{\varepsilon}(y) = X \cap B_{\varepsilon}(y) \subset X \cap Y = Y$. Thus $Y$ is open in $X$. $\blacksquare$