3.5. Subspace Topology#

Let \((X, d)\) be a metric space. Let \(Y \subseteq X\). Recall from Definition 3.3 that \((Y, d)\) is a metric subspace with the distance function \(d\) restricted to \(Y \times Y\).

Remark 3.6

Let \(Y\) be a metric subspace of \(X\) and \(S \subseteq Y\). The interior, closure and boundary of \(S\) w.r.t. \(X\) and w.r.t. \(Y\) may be different.

If a subspace hasn’t been specified, by default, we shall assume that we are computing the interior, closure and boundary w.r.t. the metric space \(X\).

3.5.1. Open Balls#

Theorem 3.38 (Open balls in the metric subspace)

Let \((Y, d)\) be a metric subspace of \((X, d)\). Let \(B_X(p, r)\) denote an open ball of radius \(r\) at \(p \in X\). Let \(B_Y(p, r)\) denote an open ball of radius \(r\) at \(p \in Y\). Then

\[ B_Y(p, r) = B_X(p, r) \cap Y. \]

Proof. We recall that

\[ B_X(p,r) = \{y \in X \ST d(p, y) < r\}. \]

Similarly,

\[ B_Y(p,r) = \{y \in Y \ST d(p, y) < r\}. \]

We first show that \(B_Y(p, r) \subseteq B_X(p, r) \cap Y\)

  1. Let \(y \in B_Y(p, r)\).

  2. Then \(y \in Y \subseteq X\) and \(d(p, y) < r\).

  3. Hence \(y \in B_X(p, r) \cap Y\).

We now show that \(B_X(p, r) \cap Y \subseteq B_Y(p, r)\).

  1. Let \(y \in B_X(p, r) \cap Y\).

  2. Then \(y \in Y\) and \(y \in X\) and \(d(p, y) < r\).

  3. Hence \(y \in Y\) and \(d(p, y) < r\).

  4. Hence \(y \in B_Y(p, r)\).

3.5.2. Open Sets#

Theorem 3.39 (Open sets in the metric subspace)

Let \((Y,d)\) be a metric subspace of \((X,d)\). Let \(S \subseteq Y\). Then \(S\) is open in \((Y,d)\) if and only if \(S = O \cap Y\) where \(O\) is an open subset of \((X,d)\).

Proof. A subset \(S\) of \(Y\) is open in \((Y, d)\) if for every \(x \in S\), there exists an open ball \(B_Y(x, r) \subseteq S \subseteq Y\).

Assume that \(S = O \cap Y\) where \(O\) is open in \(X\).

  1. Let \(x \in S\).

  2. Then \(x \in O\) and \(x \in Y\).

  3. Since \(O\) is open in \(X\), hence there is an open ball \(B_X(x, r) \subseteq O\).

  4. Then \(B_X(x, r) \cap Y \subseteq O \cap Y = S\).

  5. By Theorem 3.38,

    \[ B_X(x, r) \cap Y = B_Y(x, r) \]

    is an open ball in the metric subspace \((Y, d)\) of radius \(r > 0\) around \(x \in Y\).

  6. Hence \(S\) is open in \((Y, d)\).

For the converse, assume that \(S\) is open in \(Y\).

  1. For every \(x \in S\), there is an open ball \(B_Y(x, r_x) \subseteq S\).

  2. Hence \(\bigcup_{x \in S} B_Y(x, r_x) \subseteq S\).

  3. Also, \(x \in B_Y(x, r_x)\) implies that \(S \subseteq \bigcup_{x \in S} B_Y(x, r_x)\).

  4. Thus, \(S = \bigcup_{x \in S} B_Y(x, r_x)\).

  5. By Theorem 3.38

    \[ B_Y(x, r_x) = B_X(x, r_x) \cap Y \]

    for every \(x \in S\).

  6. Define \(T = \bigcup_{x \in S}B_X(x, r_x)\).

  7. Then \(T\) is a union of open balls of \(X\).

  8. Hence \(T\) is open in \(X\).

  9. Also

    \[\begin{split} S &= \bigcup_{x \in S} B_Y(x, r_x) \\ &= \bigcup_{x \in S}(B_X(x, r_x) \cap Y) \\ &= (\bigcup_{x \in S} B_X(x, r_x)) \cap Y \\ &= T \cap Y. \end{split}\]

  10. Hence \(S = T \cap Y\) where \(T\) is an open set of \((X, d)\).

Example 3.13

  • \([0, 1)\) is open in the metric space \(\RR_+\).

3.5.3. Subspace Topology#

Recall from Definition 3.7 that a topology on a set \(S\) is a collection of sets \(T\) such that

  1. Empty set and the whole set are elements of \(T\).

  2. \(T\) is closed under arbitrary union.

  3. \(T\) is closed under finite intersection.

Theorem 3.40 (The subspace topology)

Let \((Y, d)\) be a metric subspace of \((X, d)\). The open sets of the \((Y, d)\) form a topology.

  1. \(\EmptySet\) is open in \((X, d)\).

  2. Hence \(\EmptySet \cap Y = \EmptySet\) is open in \((Y, d)\).

  3. \(X\) is open in \((X, d)\).

  4. Hence \(X \cap Y = Y\) is open in \((Y, d)\).

  5. Let \(\{ A_i \ST i \in I \}\) be a family of open sets of \((Y, d)\).

  6. Then \(A_i = B_i \cap Y\) for every \(i \in I\) with \(B_i\) open in \(X\).

  7. Let \(B = \bigcup B_i\). Then \(B\) is open in \((X, d)\).

  8. But \(\bigcup A_i = \bigcup( B_i \cap Y) = (\bigcup B_i) \cap Y = B \cap Y\).

  9. Since \(B\) is open in \((X, d)\), hence \(B \cap Y\) is open in \((Y, d)\).

  10. Let \(\{ A_1, \dots, A_n \}\) be a finite collection of open sets of \((Y, d)\).

  11. Then \(A_i = B_i \cap Y\) for every \(i \in [1,\dots,n]\) such that \(B_i\) is open in \((X, d)\).

  12. Then \(\bigcap A_i = \bigcap (B_i \cap Y) = (\bigcap B_i) \cap Y\).

  13. Since \(\bigcap B_i\) is open in \((X, d)\) (a finite intersection), hence \(\bigcap A_i\) is open in \((Y, d)\).

3.5.4. Closed Sets#

Theorem 3.41 (Closed sets in subspace topology)

Let \(Y\) be a metric subspace of \(X\). Let \(S \subseteq Y\).

\(S\) is closed in \(Y\) if and only if \(S = C \cap Y\) where \(C\) is a closed subset of \(X\).

Proof. Let \(S\) be closed in \(Y\).

  1. Then, \(Y \setminus S\) is open in \(Y\).

  2. By definition of subspace topology,

    \[ Y \setminus S = Y \cap O \]

    where \(O\) is some open subset of \(X\).

  3. Then,

    \[\begin{split} S &= Y \setminus (Y \setminus S) \\ &= Y \setminus (Y \cap O) \\ &= Y \setminus O \\ &= Y \cap (X \setminus O). \end{split}\]

  4. But \(C = X \setminus O\) is closed in \(X\).