Proof. Since \(x_n\in A\) for infinitely many \(n\), we can choose indices \(n_1<n_2<n_3<\cdots \) with \(x_{n_k}\in A\) for every \(k\). The subsequence \(\{x_{n_k}\}\) lies in \(A\) and converges to \(x\), because every subsequence of a convergent sequence converges to the same limit. Since \(A\) is closed, \(x\in A\). □

Proof.

1.

This was noted in the definition and in the example above.

2.

Suppose \(x_n \in A\cup B\) for every \(n\) and \(x_n \longrightarrow x\). Then \(x_n\in A\) for infinitely many \(n\) or \(x_n\in B\) for infinitely many \(n\) (or both). In the first case \(x\in A\) by Lemma 1.2.3, and in the second case \(x \in B\) by the same lemma. Either way \(x\in A\cup B\), so \(A\cup B\) is closed.

3.

Suppose \(x_n \in S\) for every \(n\) and \(x_n \longrightarrow x\). For each \(i\in I\) the sequence lies in \(S_i\), and \(S_i\) is closed, so \(x\in S_i\). Since this holds for every \(i\in I\), \(x\in \bigcap _{i \in I} S_i = S\). Hence \(S\) is closed.

Proof. Let \(a\leq b\). Then \([a,b]= (-\infty , b]\cap [a,\infty )\), and \((-\infty , b]\) and \([a,\infty )\) are closed sets. Therefore, by the third part of Theorem 1.2.4, \([a,b]\) is a closed set. □

Proof. This follows by induction from the second part of Theorem 1.2.4.

The union of an infinite family of closed sets need not be closed. For example, if \(A_n = \left [ 1 + \frac {1}{n}, 5 - \frac {1}{n}\right ]\) for \(n \in \mathbb {N}\), then \(\bigcup _{n\in \mathbb {N}} A_n = (1,5)\), which is not closed: the sequence \(1+\frac {1}{n}\) lies in \((1,5)\) for \(n\geq 2\) but its limit \(1\) does not. □

Proof. We can write \[A = \{a_1\} \cup \{ a_2\}\cup \cdots \cup \{ a_r\},\] and each singleton set \(\{a_i\}\) is closed. A finite union of closed sets is closed, so \(A\) is closed. □

Proof. If every set \((x - \varepsilon , x + \varepsilon ) \cap A\) is infinite, then each contains a point of \(A\) other than \(x\), so \(x\) is a limit point of \(A\).

Conversely, let \(x\) be a limit point of \(A\) and suppose that, for some \(\varepsilon > 0\), the set \((x - \varepsilon , x + \varepsilon ) \cap A\) is finite. Its points other than \(x\), say \(y_1, \dots , y_k\), are finitely many, and there is at least one of them because \(x\) is a limit point. Put \[\varepsilon ' = \min \{|y_1 - x|, \dots , |y_k - x|\} > 0 .\] Then \((x - \varepsilon ', x + \varepsilon ')\) contains no point of \(A\) other than \(x\), contradicting the fact that \(x\) is a limit point of \(A\). Hence every set \((x - \varepsilon , x + \varepsilon ) \cap A\) is infinite. □

Proof. Suppose \(x\) is a limit point of \(A\). Taking \(\varepsilon = \frac {1}{n}\) in the definition, for every \(n \in \mathbb {N}\) there exists \(x_n \in A\) such that \(0 < |x_n - x| < \frac {1}{n}\). Then \(x_n\neq x\), and \(x_n \longrightarrow x\) because \(\frac {1}{n}\longrightarrow 0\).

Conversely, suppose \(\{ x_n \}\) is a sequence in \(A\) with \(x_n\neq x\) and \(x_n \longrightarrow x\). Let \(\varepsilon > 0\) be given. There exists \(N \in \mathbb {N}\) such that \(| x_n - x| < \varepsilon \) for every \(n > N\), so \(y = x_{N+1}\) is a point of \(A\) with \(0<|y-x|<\varepsilon \). Hence \(x\) is a limit point of \(A\). □

Proof. We prove (iv) first, since the other parts use it.

(iv)

Let \(x\in \overline {A}\). If \(x\in A\), take \(y = x\). Otherwise \(x\in A'\), and the definition of a limit point gives \(y\in A\) with \(|x-y|<\varepsilon \). Conversely, suppose every \(\varepsilon >0\) has such a \(y\). If \(x\in A\) then \(x\in \overline {A}\). If \(x\notin A\), then each such \(y\) differs from \(x\), so \(0<|x-y|<\varepsilon \) and \(x\in A'\subseteq \overline {A}\).

Taking \(\varepsilon = \frac {1}{n}\), part (iv) says that \(x \in \overline {A}\) if and only if \(x\) is the limit of some sequence \(\{y_n\}\) in \(A\).

(i)

This is immediate from \(\overline {A} = A \cup A'\).

(ii)

Let \(x\in \overline {A}\). By the remark after (iv), \(x\) is the limit of a sequence in \(A\), hence of a sequence in \(B\). Since \(B\) is closed, \(x\in B\). Thus \(\overline {A}\subseteq B\).

(v)

Suppose \(x_n\in \overline {A}\) for every \(n\) and \(x_n\longrightarrow x\). Let \(\varepsilon >0\). Choose \(n\) with \(|x_n - x|<\frac {\varepsilon }{2}\), and then, by (iv), \(y\in A\) with \(|x_n - y|<\frac {\varepsilon }{2}\). By the triangle inequality, \[|x - y| \leq |x - x_n| + |x_n - y| < \frac {\varepsilon }{2} + \frac {\varepsilon }{2} = \varepsilon .\] Hence, by (iv), \(x\in \overline {A}\), and \(\overline {A}\) is closed.

(iii)

If \(A\) is closed, then (ii) with \(B=A\) gives \(\overline {A}\subseteq A\), and with (i) this gives \(\overline {A}=A\). Conversely, if \(\overline {A}=A\) then \(A\) is closed by (v). □

Proof. Since \(S\) is bounded, \(S\subseteq [h,k]\) for some real numbers \(h<k\). Let \[T = \{t\in \mathbb {R}: S \cap (-\infty , t) \text { is finite}\}.\] Then \(h \in T\), since \(S\cap (-\infty ,h) = \emptyset \). If \(t > k\) then \(S \cap (-\infty , t) = S\) is infinite, so \(t\notin T\); hence \(k\) is an upper bound of \(T\). Note also that if \(t\in T\) and \(s<t\) then \(s\in T\). By the completeness axiom \(T\) has a supremum; let \(\xi = \sup T\).

Let \(\varepsilon > 0\). Since \(\xi -\varepsilon \) is not an upper bound of \(T\), there is \(t\in T\) with \(t>\xi -\varepsilon \), and so \(S\cap (-\infty ,\xi -\varepsilon ]\subseteq S\cap (-\infty ,t)\) is finite. On the other hand \(\xi + \varepsilon \notin T\), so \(S\cap (-\infty , \xi +\varepsilon )\) is infinite. Therefore \[S \cap (\xi - \varepsilon , \xi + \varepsilon ) = \bigl (S\cap (-\infty , \xi +\varepsilon )\bigr ) \setminus \bigl (S\cap (-\infty ,\xi -\varepsilon ]\bigr )\] is infinite. Since \(\varepsilon >0\) was arbitrary, \(\xi \) is a limit point of \(S\) by Theorem 1.2.11. □

Solution. The sequence \(\frac {1}{n}\) lies in \(A\), has all terms different from \(0\), and converges to \(0\), so \(0\in A'\) by Theorem 1.2.13.

No other point is a limit point. If \(x<0\), the neighbourhood \((x - |x|, 0)\) misses \(A\). If \(x>1\), the neighbourhood \((1,2x)\) misses \(A\). If \(0<x\leq 1\) and \(x\notin A\), then \(x\) lies strictly between two consecutive points \(\frac {1}{n+1}<x<\frac {1}{n}\), and the interval \(\left (\frac {1}{n+1},\frac {1}{n}\right )\) contains no point of \(A\). If \(x = \frac {1}{n}\in A\), then \(\left (\frac {1}{n+1},\frac {1}{n-1}\right )\) (or \(\left (\frac {1}{2},2\right )\) when \(n=1\)) contains no point of \(A\) other than \(x\).

Hence \(A' = \{0\}\) and \(\overline {A} = A\cup \{0\}\). Since \(0\in \overline {A}\setminus A\), \(A\) is not closed, but \(\overline {A} = \{0\}\cup \left \{\frac {1}{n}\right \}\) is closed.

Solution. Let \(\{x_n\}\) be a sequence of integers converging to \(x\). Taking \(\varepsilon = \frac {1}{2}\), there is \(N\) with \(|x_n - x|<\frac {1}{2}\) for \(n>N\), so \(|x_n - x_m|<1\) for \(m,n>N\). Two integers less than \(1\) apart are equal, so \(x_n\) is constant for \(n>N\), and its limit \(x\) is that integer. Hence \(\mathbb {Z}\) is closed.

For any real \(x\), the interval \(\left (x-\frac {1}{2}, x+\frac {1}{2}\right )\) contains at most one integer, hence only finitely many. By Theorem 1.2.11, \(x\) is not a limit point of \(\mathbb {Z}\). So \(\mathbb {Z}' = \emptyset \).

Solution. (a) Every \(x\in [0,1]\) is a limit point of \((0,1]\): if \(0\leq x<1\) the points \(x + \frac {1-x}{n+1}\) lie in \(A\), differ from \(x\) and converge to \(x\), and for \(x=1\) use \(1-\frac {1}{n+1}\). A point outside \([0,1]\) has a neighbourhood missing \([0,1]\). So \(A' = [0,1] = \overline {A}\).

(b) Every interval contains rational numbers, so every neighbourhood of a point \(x\in [0,1]\) contains rationals of \([0,1]\) other than \(x\). Hence \(B' = [0,1]\) and \(\overline {B} = [0,1]\), although \(B\) itself omits every irrational point of \([0,1]\).