Interior of a set
Definition (Interior point and interior of a set)
Let \((E, d)\) be a metric space and \(A \subseteq E\). We say that \(x_{0} \in A\) is an interior point of \(A\) if there exists \(r > 0\) with \(B(x_{0}, r) \subseteq A\). The interior of \(A\), denoted \(A^{\circ}\), is the set of interior points of \(A\).
Lemma (\(A^{\circ}\) is the largest open set contained in \(A\))
Let \(G \subseteq A\) with \(G\) open. Then \(G \subseteq A^{\circ}\). In particular, \(A^{\circ}\) is the largest open set contained in \(A\).
Proof: If \(x_{0} \in G\), since \(G\) is open there exists \(r > 0\) with \(B(x_{0}, r) \subseteq G \subseteq A\), so that \(x_{0} \in A^{\circ}\). We conclude that \(G \subseteq A^{\circ}\).
Exercise (\(A^{\circ}\) is open)
Prove that for every \(A \subseteq E\), \(A^{\circ}\) is an open set.
Lemma (Interior of an intersection)
Let \(A_{1}, A_{2} \subseteq E\). Then \((A_{1} \cap A_{2})^{\circ} = A_{1}^{\circ} \cap A_{2}^{\circ}\).
Proof: Since \(A_{1}^{\circ} \cap A_{2}^{\circ}\) is open and is contained in \(A_{1} \cap A_{2}\), by the previous lemma \(A_{1}^{\circ} \cap A_{2}^{\circ} \subseteq (A_{1} \cap A_{2})^{\circ}\).
Conversely, if \(x \in (A_{1} \cap A_{2})^{\circ}\), there exists \(r > 0\) with \(B(x, r) \subseteq A_{1} \cap A_{2}\). Then \(B(x, r) \subseteq A_{1}\) and \(B(x, r) \subseteq A_{2}\), which implies \(x \in A_{1}^{\circ} \cap A_{2}^{\circ}\).
Exercise (Monotonicity of the interior)
If \(A \subseteq B\) then \(A^{\circ} \subseteq B^{\circ}\). Hint: use the lemma \(G \subseteq A^{\circ}\) whenever \(G\) is open and \(G \subseteq A\).
Exercise (Characterisation of open sets by the interior)
Given \(B \subseteq E\), \(B\) is open if and only if \(B = B^{\circ}\).
Convergence of sequences and distance between sets
Definition (Convergence of sequences)
A sequence \(\{ x_{n} \}_{n=1}^{\infty} \subseteq E\) converges to \(x \in E\) if for every \(\varepsilon > 0\) there exists \(n_{0} \in \mathbb{N}\) such that \(n \geq n_{0}\) implies \(d(x_{n}, x) < \varepsilon\). We write \(x_{n} \to x\) as \(n \to \infty\).
Definition (Distance between sets)
Given \(A, B \subseteq E\), we define
\[d(A, B) = \inf \{ d(x, y) : (x, y) \in A \times B \}.\]In particular, for \(x \in E\) and \(A \subseteq E\), \(d(x, A) = \inf\{ d(x,a) : a \in A\}\).
Exercise (The distance to a set is \(1\)-Lipschitz)
| For every \(A \subseteq E\) and \(x, y \in E\), $$ | d(x, A) - d(y, A) | \leq d(x, y)$$. |
Closure of a set
Lemma (Characterisation of the closure by sequences and by distance)
Given \(x \in E\) and \(A \subseteq E\), the following are equivalent:
- \(d(x, A) = 0\);
- there exists \(\{ x_{n} \}_{n=1}^{\infty} \subseteq A\) such that \(x_{n} \to x\) as \(n \to \infty\).
Proof: \((1) \implies (2)\): If \(d(x, A) = 0\), for each \(n \in \mathbb{N}\) there exists \(x_{n} \in A\) with \(d(x, x_{n}) < 1/n\), so that \(\{x_{n}\}_{n=1}^{\infty} \subseteq A\) and \(x_{n} \to x\).
\((2) \implies (1)\): If \(\{x_{n}\}_{n=1}^{\infty} \subseteq A\) and \(x_{n} \to x\), then \(d(x, A) \leq d(x, x_{n}) \to 0\), whence \(d(x, A) = 0\).
Definition (Closure and adherent points)
Given \(A \subseteq E\), the closure of \(A\) is
\[\overline{A} = \{ x \in E : d(x, A) = 0\}.\]Its elements are called adherent points of \(A\). It is immediate that \(A \subseteq \overline{A}\).
Proposition (Monotonicity of the closure)
If \(A \subseteq B \subseteq E\), then \(\overline{A} \subseteq \overline{B}\).
Proof: For \(x \in E\),
\[\inf_{b \in B} d(x, b) \leq \inf_{a \in A} d(x, a),\]i.e. \(d(x, B) \leq d(x, A)\). Thus, \(d(x, A) = 0\) implies \(d(x, B) = 0\).
Example (Zero distance between two disjoint closed sets)
Let \(A = \{ n + \tfrac{1}{n} \}_{n=1}^{\infty}\) and \(B = \mathbb{Z}\). Both are closed (for example \(\mathbb{R} \setminus B = \bigcup_{n \in \mathbb{Z}} (n, n+1)\) is open). Nevertheless,
\[d\!\left(n + \tfrac{1}{n}, n\right) = \tfrac{1}{n} \to 0,\]so that \(d(A, B) = 0\).
Lemma (Closed is equivalent to being equal to its closure)
Let \(A \subseteq E\). Then \(A\) is closed if and only if \(A = \overline{A}\).
Proof: \((\impliedby)\): If \(A = \overline{A}\) and \(x \notin A\), then \(d(x, A) > 0\). There exists \(r > 0\) with \(r \leq d(x, A)\) and then \(B(x, r) \cap A = \emptyset\), i.e. \(B(x, r) \subseteq E \setminus A\). Thus \(E\setminus A\) is open and \(A\) is closed.
\((\implies)\): Suppose \(A\) is closed and take \(x \in E \setminus A\). Since \(E\setminus A\) is open, there exists \(r_{1} > 0\) with \(B(x, r_{1}) \subseteq E \setminus A\), so that \(B(x, r_{1}) \cap A = \emptyset\) and \(d(x, A) \geq r_{1} > 0\). Hence \(x \notin \overline{A}\), which proves \(\overline{A} \subseteq A\). The inclusion \(A \subseteq \overline{A}\) is immediate.
Proposition (The closure is a closed set)
For every \(A \subseteq E\), \(\overline{A}\) is closed.
Proof: Let \(z \notin \overline{A}\), that is \(d(z, A) = r > 0\). We shall see that \(B(z, r/2) \subseteq E \setminus \overline{A}\). If \(w \in B(z, r/2)\) and \(a \in A\), then
\[r \leq d(z, a) \leq d(z, w) + d(w, a) < \tfrac{r}{2} + d(w, a),\]so that \(d(w, a) > r/2\) and, taking the infimum, \(d(w, A) \geq r/2 > 0\), i.e. \(w \notin \overline{A}\).
Exercise (Properties of the closure under unions and intersections)
For \(A, B \subseteq E\):
- \(\overline{A \cup B} = \overline{A} \cup \overline{B}\);
- \(\overline{A \cap B} \subseteq \overline{A} \cap \overline{B}\).
Accumulation and boundary
Definition (Accumulation point)
A point \(x_{0} \in E\) is an accumulation point of \(A \subseteq E\) if for every \(r > 0\),
\[\big(B(x_{0}, r) \setminus \{ x_{0}\}\big) \cap A \neq \emptyset.\]That is, every ball centred at \(x_{0}\) contains points of \(A\) other than \(x_{0}\).
Exercise (Characterisation of accumulation points by sequences)
\(x_{0}\) is an accumulation point of \(A\) if and only if there exists a sequence \(\{ x_{n}\}_{n=1}^{\infty} \subseteq A \setminus \{x_{0}\}\) such that \(x_{n} \to x_{0}\).
Example (Unique accumulation point of \(\{1/n\}\))
Let \(A = \{ 1/n : n \geq 1\}\). Then \(\overline{A} = A \cup \{ 0\}\) and \(0\) is the only accumulation point of \(A\), since
\[B\!\left(\tfrac{1}{n}, \tfrac{1}{(n+1)^{2}}\right) \cap A = \left\{ \tfrac{1}{n}\right\} \quad \text{if } n \geq 1.\]Definition (Boundary point and boundary of a set)
Given \(A \subseteq E\) and \(x \in E\), we say that \(x\) is a boundary point of \(A\) if for every \(r > 0\),
\[B(x, r) \cap A \neq \emptyset \quad\text{and}\quad B(x, r) \cap (E \setminus A) \neq \emptyset.\]Equivalently, \(d(x, A) = d(x, E \setminus A) = 0\).
Notation (Boundary)
The boundary of \(A\) is denoted
\[\partial A = \{\text{boundary points of } A\} = \{ x \in E : d(x, A) = d(x, E \setminus A) = 0\} = \overline{A} \cap \overline{E \setminus A}.\]Note that \(\partial A \subseteq \overline{A}\).
Neighbourhoods and density
Definition (Neighbourhood of a point)
Given \(x \in E\), \(V \subseteq E\) is a neighbourhood of \(x\) if there exists \(r > 0\) such that \(B(x, r) \subseteq V\).
Note (Open sets as neighbourhoods)
If \(V\) is open, \(V\) is a neighbourhood of each of its points.
Definition (\(r\) neighbourhood of a set)
Given \(A \subseteq E\) and \(r > 0\), we define
\[V_{r}(A) = \{ x \in E : d(x, A) < r\}.\]In particular \(A \subseteq \overline{A} \subseteq V_{r}(A)\).
Exercise (Balls centred in \(A\) are contained in \(V_{r}(A)\))
Let \(x \notin \overline{A}\) with \(x \in V_{r}(A)\). If \(0 < r_{1} < r - d(x, A)\), show that
\[B(x, r_{1}) \subseteq V_{r}(A).\]Lemma (\(V_{r}(A)\) is open)
If \(A \subseteq E\) and \(r > 0\), then \(V_{r}(A)\) is open.
Proof: Let \(x \in V_{r}(A)\) and take \(a \in A\) with \(d(x, a) < r\). Define \(r_{1} = r - d(x, a) > 0\). For \(y \in B(x, r_{1})\) we have that
\[d(y, A) \leq d(y, a) \leq d(y, x) + d(x, a) < r_{1} + d(x, a) = r,\]that is, \(y \in V_{r}(A)\). Thus \(B(x, r_{1}) \subseteq V_{r}(A)\).
Proposition (The closure is an intersection of neighbourhoods)
For every \(A \subseteq E\),
\[\overline{A} = \bigcap_{r > 0} V_{r}(A).\]In particular, \(\overline{A}\) is a countable intersection of open sets.
Proof: If \(x \in \overline{A}\) then \(d(x, A) = 0 < r\) for every \(r > 0\), so \(x \in V_{r}(A)\). Conversely, if \(x \in \bigcap_{r > 0} V_{r}(A)\) then \(d(x, A) < r\) for every \(r > 0\), so that \(d(x, A) = 0\) and \(x \in \overline{A}\).
Definition (Dense set and separable space)
\(A \subseteq E\) is dense in \(E\) if \(\overline{A} = E\). A metric space \((E, d)\) is separable if it possesses a countable dense set.
Example (\(\mathbb{R}\) and \(\mathbb{R}^{d}\) are separable)
| $$(\mathbb{R}, | \cdot | )\(is separable since\)\mathbb{Q}\(is dense in\)\mathbb{R}\(. In general\)(\mathbb{R}^{d}, |\cdot|)\(is separable because\)\mathbb{Q}^{d}\(is dense in\)\mathbb{R}^{d}$$. |