Semantics

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2.1.  Metric Spaces

Definition 2.1.1. Let \(P\) and \(R\) be a point sort and a scalar sort, respectively. Then, we let

\[\mathcal L_{\mathrm{met}} = \left\{ P,R,\, 0_R,+_R,<_{R},\, d \right\},\]

where \(0_R\) is a constant of sort \(R, +_R:R^2\to R, <_{R}\subseteq R^2,\) and \(d:P^2\to R\) is a function symbol. An \(\mathcal L_{\mathrm{met}}\)-structure therefore has the form

\[\mathcal M = \left( M,S;\, 0^{\mathcal M},+^{\mathcal M},<^{\mathcal M},d^{\mathcal M} \right),\]

where \(M\) is the set of points and \(S\) is the set in which distances take their values. We henceforth suppose that the scalar sort is interpreted as the ordered additive structure of the real numbers. That is,

\[\mathcal M_{(M,d)} = (M,\mathbb R;\,0,+,<,d).\]


Definition 2.1.2. \(\mathcal M_{(M,d)}\) is a metric space if it satisfies the following sentences, where \(x,y,z\) range over the point sort \(P\!:\)

  1. \(\forall x,y\ (0\le d(x,y)),\)
  2. \(\forall x,y\ (d(x,y)=0\leftrightarrow x=y),\)
  3. \(\forall x,y\ (d(x,y)=d(y,x)),\)
  4. \(\forall x,y,z\ \left(d(x,z)\le d(x,y)+d(y,z)\right).\)

In this case, \(d\) is called the metric on \(M.\)

Definition 2.1.3. A set \(S\subseteq \mathbb R^k\) is convex if

\[\forall x,y\in S\ \forall \lambda \in (0,1)\ \left( \lambda x + (1-\lambda)y\in S\right).\]


Example 2.1.4. Let \(k\in \mathbb N_{>0}\) and \(a_i<b_i\) for \(i=1,\dots, k,\) and define \(S\) to be the set of all points \(x=(x_1,\dots,x_k)\) in \(\mathbb R^k\) such that \(x_i\in [a_i,b_i]\) for \(i=1,\dots, k.\) The set \(S\) is called a \(k\)-cell.
 Let \(x,y\in S, \lambda \in (0,1),\) and \(z_i = \lambda x_i + (1-\lambda)y_i.\) Since

\[\lambda a_i\le \lambda x_i \le \lambda b_i\quad \text{and} \quad (1-\lambda)a_i\le (1-\lambda)y_i\le (1-\lambda)b_i,\]

we have \(a_i\le z_i\le b_i,\) so that \(z_i\in S.\) Thus \(S\) is convex.
 In general, \(k\)-cells are convex.

Definition 2.1.5. Let \((M,d)\) be a metric space, and let \(a\in M,r>0,\) and \(X\subseteq M.\)

  1. \(X\) is the open ball of radius \(r\) centered at \(a\) if and only if

    \[\forall x\in M\ (x\in X \leftrightarrow d(x,a)<r);\]

    the unique set is denoted by \(B_r(a).\)

  2. \(X\) is the closed ball of radius \(r\) centered at \(a\) if and only if

    \[\forall x\in M\ (x\in X \leftrightarrow d(x,a)\le r);\]

    the unique set is denoted by \(\overline B_r(a).\)

  3. \(X\) is the punctured open ball of radius \(r\) centered at \(a\) if and only if

    \[\forall x\in M\ (x\in X \leftrightarrow 0<d(x,a)<r);\]

    the unique set is denoted by \(B_r^{\ast}(a).\)


Definition 2.1.6. Let \((M,d)\) be a metric space, let \(a\in M,\) and let \(X\subseteq M.\) The set \(X\) is a neighborhood of \(a\) if and only if

\[\exists r>0\ (B_r(a)\subseteq X).\]


Definition 2.1.7. Let \((M,d)\) be a metric space, let \(A\subseteq M\), and let \(x\in M\). The metric induces several local relations between the point \(x\) and the set \(A\).

  1. \(x\) is an interior point of \(A\) if and only if \(\exists r>0 \ \left(B_r(x)\subseteq A\right).\)
  2. \(x\) is a boundary point of \(A\) if and only if \(\forall r>0\ \left( B_r(x)\cap A\neq\varnothing \land B_r(x)\cap A^c\neq\varnothing \right).\)
  3. \(x\) is an exterior point of \(A\) if and only if \(\exists r>0\ \left(B_r(x)\subseteq A^{c}\right).\)
  4. \(x\) is a limit point of \(A\) if and only if \(\forall r>0 \ \left(B_r^{\ast}(x)\cap A\neq\varnothing\right).\)
  5. \(x\) is a closure point of \(A\) if and only if \(\forall r>0 \ \left(B_r(x)\cap A\neq\varnothing\right).\)


Definition 2.1.8. Let \((M,d)\) be a metric space and let \(A,X\subseteq M.\) The metric induces several relations between the two sets.

  1. \(X\) is the interior of \(A\) if and only if

    \[\forall x\in M\ (x\in X \leftrightarrow x\text{ is an interior point of }A);\]

    the unique set is denoted by \(\operatorname{int} A.\)

  2. \(X\) is the boundary of \(A\) if and only if

    \[\forall x\in M\ (x\in X \leftrightarrow x\text{ is a boundary point of }A);\]

    the unique set is denoted by \(\partial A.\)

  3. \(X\) is the exterior of \(A\) if and only if

    \[\forall x\in M\ (x\in X \leftrightarrow x\text{ is an exterior point of }A);\]

    the unique set is denoted by \(\operatorname{ext} A.\)

  4. \(X\) is the derived set of \(A\) if and only if

    \[\forall x\in M\ (x\in X \leftrightarrow x\text{ is a limit point of }A);\]

    the unique set is denoted by \(A'.\)

  5. \(X\) is the closure of \(A\) if and only if

    \[\forall x\in M\ (x\in X \leftrightarrow x\text{ is a closure point of }A);\]

    the unique set is denoted by \(\overline A.\)


Theorem 2.1.9. Let \((M,d)\) be a metric space and let \(A\subseteq M.\) Every point of \(M\) is either an interior point of \(A,\) an exterior point of \(A,\) or a boundary point of \(A,\) and hence

\[M = \operatorname{int}A \ \dot{\cup} \ \partial A\ \dot{\cup} \ \operatorname{ext}A.\]

Proof. If \(x\in \operatorname {int} A,\) then \(B_r(x)\subseteq A\) for some \(r>0.\) Since \(x\in \operatorname{int} A\) implies \(x\in A,\) \(B_r(x)\) does not meet \(A^{c}.\) Hence \(\operatorname{int} A \cap \partial A = \varnothing.\) If \(0<s<r\) then \(B_s(x)\subseteq B_r(x)\subseteq A.\) If \(s>r\) then \(B_r(x)\subseteq B_s(x)\) so that \(B_s(x)\cap A\ne \varnothing.\) Hence \(\operatorname{int} A \cap \operatorname{ext} A = \varnothing.\) Lastly, since \(x\in \partial A\) implies \(x\notin \operatorname{ext} A,\) \(\partial A \cap \operatorname{ext} A = \varnothing.\) Therefore \(\operatorname{int} A, \partial A,\) and \(\operatorname{ext} A\) are mutually disjoint.
 We claim that \(\partial A \cup \operatorname{ext} A = \left(\operatorname{int} A\right)^{c}.\) Indeed, if \(x\in \partial A\) then \(B_r(x)\cap A^{c}\ne \varnothing\); hence \(x\notin \operatorname{int} A.\) If \(x\in \operatorname{ext} A\) then \(B_r(x)\subseteq A^{c}\) for some \(r>0\); hence \(x\notin \operatorname{int} A.\) Therefore \(\partial A \cup \operatorname{ext} A \subseteq \left( \operatorname{int} A\right) ^{c}.\) Conversely, suppose \(x\notin \operatorname{int} A,\) which means that every ball around \(x\) meets \(A^{c}.\) There may exist some \(r>0\) such that \(B_r(x)\subseteq A^c,\) which means \(x\in \operatorname{ext} A.\) Otherwise, no such \(r\) exists, which means that every ball around \(x\) meets \(A.\) Thus \(x\in \partial A.\) Therefore \(\left(\operatorname{int} A\right)^{c}\subseteq \partial A \cup \operatorname{ext} A.\) It is concluded that \(\partial A \cup \operatorname{ext} A = \left(\operatorname{int} A\right)^{c}.\) \(\square\)

Proposition 2.1.10. Let \((M,d)\) be a metric space and let \(A\subseteq M.\) Then the following statements hold.

  1. \(\overline A = M \setminus \operatorname{ext} A = \operatorname{int} A \cup \partial A.\)
  2. \(\overline A = A\cup A'.\)
  3. \(\operatorname{int} A \subseteq A \subseteq \overline A.\)

Proof. (1) If \(x\in \overline A,\) then every ball meets \(A.\) Hence no balls lie entirely in \(A^{c},\) so that \(x\notin \operatorname{ext} A.\) If \(x\notin \operatorname{ext} A,\) then every ball meets \(A\) so that \(x\in \overline A.\) Hence \(\overline A = M \setminus \operatorname{ext} A.\)
 (2) Since \(B_r(x) = B_r^{\ast}(x) \cup \lbrace x\rbrace,\) to say that every ball around \(x\) meets \(A\) is equivalent to say that every punctured ball meets \(A\) or \(\lbrace x\rbrace\) meets \(A.\) Hence \(\overline A = A\cup A'.\)
 (3) If \(x\in \operatorname{int} A,\) then \(B_r(x)\subseteq A\) for some \(r>0.\) Since \(x\in B_r(x),\) \(x\in A,\) so that \(\operatorname{int} A \subseteq A.\) (2) immediately yields \(A \subseteq \overline A.\) Therefore \(\operatorname{int} A \subseteq A \subseteq \overline A.\)\(\square\)

Definition 2.1.11. Let \((M,d)\) be a metric space and let \(A\subseteq M\).

  1. \(A\) is open if and only if \(A\subseteq \operatorname{int} A.\)
  2. \(A\) is closed if and only if \(A'\subseteq A.\)
  3. \(A\) is perfect if and only if \(A=A'.\)
  4. \(A\) is dense in \(M\) if and only if \(M\subseteq \overline A.\)
  5. \(A\) is bounded if and only if \(A\subseteq B_r(x)\) for some \(x\in M\) and \(r>0.\)


Proposition 2.1.12. Let \((M,d)\) be a metric space and let \(A\subseteq M\). Then the following statements hold.

  1. \(A\) is open if and only if \(A\cap\partial A=\varnothing.\)
  2. \(A\) is closed if and only if \(\partial A\subseteq A.\)
  3. \(A\) is dense in \(M\) if and only if \(\operatorname{ext} A = \varnothing.\)
  4. \(A\) is bounded if and only if \(\overline A\) is bounded.

Proof. (1) Suppose \(A\) is open. Since \(\operatorname{int} A\) and \(\partial A\) is disjoint, \(A\subseteq \operatorname{int} A\) implies \(A\cap \partial A = \varnothing.\) Conversely, suppose \(A\cap \partial A = \varnothing.\) Since \(x\in A\) implies \(x\notin \partial A,\) we have then \(x\in \operatorname{int} A\) or \(x\in \operatorname{ext} A.\) Since \(A\) and \(\operatorname{ext} A\) are disjoint, \(x\in \operatorname{int} A.\) Hence \(A\subseteq \operatorname{int} A,\) so \(A\) is open.
 (2) If \(\partial A\subseteq A,\) then \(\operatorname{int} A \cup \partial A \subseteq A.\) Hence \(A'\subseteq \overline A \subseteq A.\) Thus \(A\) is closed. Conversely, if \(A'\subseteq A\) then \(\overline A\subseteq A.\) Hence \(\partial A \subseteq \overline A \subseteq A.\)
 (3) Since \(M\subseteq M\setminus \operatorname{ext} A\) is equivalent to \(\operatorname{ext} A=\varnothing,\) if \(A\) is dense in \(M\) then \(\operatorname{ext} A=\varnothing,\) and vice versa.
 (4) Suppose that \(A\) is bounded, and choose \(r>0\) and \(a\in M\) such that \(A\subseteq B_r(a).\) If \(x\in A'\) then \(B_1^{\ast}(x)\) contains some point \(y\in A.\) Then \(d(x,y)<1,\) while \(d(y,a)<r,\) so the triangle inequality gives \(d(x,a)<r+1.\) This proves that \(A'\subseteq B_{r+1}(a),\) which is enough for boundedness. Conversely, if \(\overline A\) is bounded, then \(\overline A = A \cup A' \subseteq B_r(a)\) for some \(a\in M\) and \(r>0.\) Thus \(A\subseteq B_r(a),\) so \(A\) is bounded.\(\square\)

Proposition 2.1.13. Let \((M,d)\) be a metric space and let \(A\subseteq M\). Then the following statements hold.

  1. If \(A\) is both closed and dense in \(M\), then \(A=M.\)
  2. If \(A\) is bounded, then \(\operatorname{int} A, \partial A,\) and \(A'\) are bounded.
  3. If \(A\) is both bounded and dense in \(M\), then \(M\) is bounded. Consequently, if \(M\) is unbounded, every dense subset of \(M\) is unbounded.

Proof. (1) Since \(A\) is closed, \(\overline A = A.\) Thus, \(A\)’s being dense in \(M\) implies \(M\subseteq A.\) Therefore \(A=M.\)
 (2) Since \(A\) is bounded, \(\overline A\) is also bounded by (4) of Proposition 2.1.12. Hence \(\overline A\subseteq B_r(a)\) for some \(r>0\) and \(a\in M.\) Since \(\operatorname{int} A, \partial A,\) and \(A'\) are subsets of \(\overline A,\) the sets are also subsets of \(B_r(a).\) Therefore \(\operatorname{int} A, \partial A,\) and \(A'\) are bounded.
 (3) If \(A\) is bounded, then \(\overline A\) is also bounded, so that \(\overline A\subseteq B_r(a)\) for some \(r>0\) and \(a\in M.\) Since \(A\) is dense in \(M,\) it follows that \(M\subseteq \overline A \subseteq B_r(a).\) Hence \(M\) is bounded.\(\square\)

Theorem 2.1.14. Every open ball is open.

Proof. Let \((M,d)\) be a metric space. Then let \(a\in M\) and \(r>0\) be arbitrary, and let \(A=B_r(a).\) We show that \(x\in A\) implies \(x\in \operatorname {int} A.\) If \(x\in A\) then \(d(x,a)<r.\) We choose \(s>0\) so that \(s<r-d(x,a).\) Since, for any \(y\in M,\)

\[d(y,x)<s \quad \Longrightarrow \quad d(y,a)\le d(y,x)+d(x,a)<s+d(x,a)<r,\]

we have \(B_s(x)\subseteq A.\) Hence \(x\in \operatorname {int} A.\) Therefore it is concluded that \(A\subseteq \operatorname {int} A\) and \(A\) is open. \(\square\)

Proposition 2.1.15. Let \((M,d)\) be a metric space, and let \(A\subseteq M,\) then

  1. \(\overline A\) is closed,
  2. \(\overline A \subseteq X\) for every closed set \(X\subseteq M\) such that \(A\subseteq X.\)

Proof. (1) Let \(x\notin \overline A.\) Since \(\overline A = A\cup A',\) there exists \(r>0\) such that \(B_r(x)\cap A = \varnothing.\) We show that there exists \(s>0\) such that \(B_s(x)\cap \overline A=\varnothing.\) Equivalently, we show that there exists \(s>0\) such that

\[y\in B_s(x) \quad \Longrightarrow \quad \exists t>0\ (B_t(y)\cap A = \varnothing).\]

Put \(s=r/2\) and \(t=r/2.\) Then, \(d(z,x)\le d(y,x)+d(z,y)<r,\) for every \(z\in B_t(y).\) Hence \(B_t(y)\cap A = \varnothing\) so that \(B_s(x)\cap \overline A=\varnothing.\) That is, \(x\) is not a limit point of \(\overline A.\) Consequently, every limit point of \(\overline A\) is in \(\overline A,\) and hence \(\overline A\) is closed.
 (2) Let \(x\) be a limit point of \(A.\) Since \(A\subseteq X,\) that \(B^{\ast}_r(x)\) meets \(A\) for all \(r>0\) implies that \(B^{\ast}_r(x)\) meets \(X\) for all \(r>0.\) Hence \(x\) is a limit point of \(X.\) Since \(X\) is closed, \(x\in X\) so that \(A'\subseteq X.\) Therefore \(\overline A = A \cup A'\subseteq X.\)\(\square\)

Theorem 2.1.16. Let \((M,d)\) be a metric space, and let \(A\subseteq M\). If \(a\in A'\), then, for every \(r>0\), the set \(B_r(a)\cap A\) is infinite.

Proof. Suppose, for contradiction, that there exists \(r>0\) such that \(B_r(a)\cap A\) is finite. Then \(B_r^{\ast}(a)\cap A\) is also finite. Since \(a\in A'\), this set is nonempty. Thus, for some \(n\in\mathbb N_{>0}\), we may write

\[B_r^{\ast}(a)\cap A=\lbrace p_1,\ldots,p_n\rbrace .\]

Define

\[s=\min_{1\leq i\leq n} d(a,p_i).\]

Since \(p_i\neq a\) for every \(i\), we have \(d(a,p_i)>0\), and hence \(s>0\). Moreover, since every \(p_i\in B_r(a)\), we have \(s<r\).
 We now claim that \(B_s^{\ast}(a)\cap A=\varnothing.\) Indeed, if \(x\in B_s^{\ast}(a)\cap A\) then we have \(x\in B_r^{\ast}(a)\cap A,\) since \(s<r.\) Thus \(x=p_i\) for some \(i,\) and

\[s \le d(a,p_i) = d(a,x),\]

which contradicts \(x\in B^{\ast}_s(a).\) Therefore \(B_s^{\ast}(a)\cap A=\varnothing\), contradicting \(a\in A'\). Hence it is concluded that \(B_r(a)\cap A\) is infinite for every \(r>0\). \(\square\)

Theorem 2.1.17. Let \((M,d)\) be a metric space. A set \(A\subseteq M\) is open if and only if \(A^c\) is closed.

Proof. Suppose that \(A\) is open and \(x\in A.\) Then we can choose \(r>0\) so that \(B_r(x)\subseteq A.\) It follows that \(B^{\ast}_r(x)\cap A^c = \varnothing,\) so \(x\) is not a limit point of \(A^c.\) Hence every limit point of \(A^c\) is a member of \(A^c,\) and thus \(A^c\) is closed.
 Conversely, suppose that \(A^c\) is closed and let \(x\in A.\) Then \(x\) is not a limit point of \(A^c.\) Hence we can choose \(r>0\) so that \(B^{\ast}_r(x)\cap A^c=\varnothing.\) It follows that \(B_r(x)\subseteq A.\) Therefore \(A\) is open.\(\square\)

Theorem 2.1.18. Let \((M,d)\) be a metric space, and let \(A_i\subseteq M\) for every \(i\in I.\)

  1. If \(A_i\) is open for every \(i\in I,\) then \(\bigcup_{i\in I } A_i\) is open.
  2. If \(A_i\) is closed for every \(i\in I,\) then \(\bigcap_{i\in I } A_i\) is closed.

Proof. (1) Let \(x\in \bigcup_{i\in I } A_i.\) Then we may let \(x\in A_i\) for some \(i\in I.\) Since \(A_i\) is open, there exists \(r>0\) such that \(B_r(x)\subseteq A_i.\) Since \(A_i\subseteq \bigcup_{i\in I } A_i,\) we have \(B_r(x)\subseteq \bigcup_{i\in I } A_i.\) Therefore \(\bigcup_{i\in I } A_i\) is open.
 (2) By Theorem 2.1.17, \(A_i^c\) is open for every \(i\in I.\) Then \(\bigcup_{i\in I } A^c_i\) is open by (1). Since

\[\bigcup_{i\in I } A^c_i = \lowparen{\bigcap_{i\in I } A_i}^{\!\!c},\]

\(\bigcap_{i\in I } A_i\) is closed by Theorem 2.1.17.\(\square\)

Theorem 2.1.19. Let \((M,d)\) be a metric space, and let \(A_i\subseteq M\) for every \(i\in I\) where \(I\) is a finite set.

  1. If \(A_i\) is open for every \(i\in I,\) then \(\bigcap_{i\in I } A_i\) is open.
  2. If \(A_i\) is closed for every \(i\in I,\) then \(\bigcup_{i\in I } A_i\) is closed.

Proof. (1) Let \(x\in \bigcap_{i\in I } A_i.\) Then, for every \(i\in I,\) we have \(x\in A_i.\) Since \(A_i\) is open, there exists \(r_i>0\) such that \(B_{r_i}(x)\subseteq A_i.\) We let

\[r = \min_{i\in I}r_i.\]

Then \(B_r(x)\subseteq A_i\) for every \(i\in I,\) so that \(B_r(x)\subseteq \bigcap_{i\in I } A_i.\) Hence \(\bigcap_{i\in I } A_i\) is open.
 (2) By Theorem 2.1.17, \(A_i^c\) is open for every \(i\in I.\) Then \(\bigcap_{i\in I } A^c_i\) is open by (1). Since

\[\bigcap_{i\in I } A^c_i = \lowparen{\bigcup_{i\in I } A_i}^{\!\!c},\]

\(\bigcup_{i\in I } A_i\) is closed by Theorem 2.1.17.\(\square\)

Remark 2.1.20. The necessity of \(I\)’s being finite in the preceding theorem can be illustrated by the following example. For \(n\in \mathbb N_{>0},\) we let

\[A_n = \left( -\frac 1 n, \frac 1 n\right).\]

Then \(A_n\) is open in \(\mathbb R.\) However \(\bigcap_{n\in \mathbb N_{>0}}A_n\) consists of a single point, namely, \(x=0,\) and is therefore not open in \(\mathbb R.\) Similarly, the union of an infinite set of closed sets need not be closed.

Theorem 2.1.21. Let \(A\) be a nonempty subset of \(\mathbb R\) which is bounded above, and let \(\alpha = \sup A.\) Then \(\alpha \in \overline A,\) and thus \(\alpha \in A\) if \(A\) is closed.

Proof. Since \(\alpha\) is the supremum for \(A,\) for every \(r>0,\) there exists \(x\in A\) such that \(0\le \alpha -x <r.\) Hence \(\alpha \in A\) or \(\alpha\in A'.\) Hence \(\alpha \in \overline A.\)\(\square\)

Definition 2.1.22. Let \((M,d)\) be a metric space and let \(A\subseteq E \subseteq M.\) \(A\) is open relative to \(E\) if and only if

\[\forall x \in A\ \exists r>0\ (B_r(x)\cap E \subseteq A).\]


Theorem 2.1.23. Let \((M,d)\) be a metric space and let \(A\subseteq E\subseteq M.\) Then, \(A\) is open relative to \(E\) if and only if there exists an open set \(X\subseteq M\) such that \(A = X\cap E.\)

Proof. Suppose that \(A\) is open relative to \(E.\) Let \(x\in A\) and define \(r_x\) to be a positive real such that \(B_{r_x}(x)\cap E \subseteq A.\) Then we define

\[X = \bigcup_{x\in A}B_{r_x}(x).\]

Since every open ball is open, \(X\) is open by Theorem 2.1.18.
 Since \(x\in B_{r_x}(x)\) for all \(x,\) we have \(A\subseteq X \cap E.\) We also have \(X\cap E\subseteq A\) since \(B_{r_x}(x)\cap E \subseteq A\) for all \(x.\) Thus \(A=X\cap E.\)
 Conversely, suppose that a set \(X\) is open and \(X \cap E=A.\) Then, for every \(x\in A,\) there exists \(r>0\) such that \(B_r(x)\subseteq X.\) It follows that \(B_{r}(x)\cap E\subseteq A.\) Thus \(A\) is open relative to \(E.\)\(\square\)

References

  1. Rudin, W. (1976). Principles of mathematical analysis (3rd ed.). McGraw-Hill Education.
  2. Munkres, J. R. (2000). Topology (2nd ed.). Pearson.