Proof. Suppose \(x\notin A\). Since \(A\) is closed, \(A^{C}\) is open, so there is \(\varepsilon >0\) with \(\left (x-\varepsilon ,\,x+\varepsilon \right )\subseteq A^{C}\). But \(x_n\rightarrow x\) means \(x_n\) lies in that interval for all large \(n\), so \(x_n\notin A\) for all large \(n\) — contradicting that \(x_n\in A\) frequently, which puts terms of \(A\) arbitrarily far along the sequence. Hence \(x\in A\). □

Proof. Each part is the complement of the corresponding statement about open sets.

(i)

\(\emptyset \) is closed because its complement \(\mathbb {R}\) is open, and \(\mathbb {R}\) is closed because its complement \(\emptyset \) is open.

(ii)

If \(A\) and \(B\) are closed then \(A^{C}\) and \(B^{C}\) are open, and \[\left (A\cup B\right )^{C} = A^{C}\cap B^{C}\] is an intersection of two open sets, hence open. So \(A\cup B\) is closed.

(iii)

For any family \(\Omega =\{S_i\}\) of closed sets, \[\left (\bigcap _{i\in I} S_i\right )^{C} = \bigcup _{i\in I} S_i^{C},\] a union of open sets, which is open. Hence the intersection is closed. □

Proof.

1.

\(\emptyset \) is closed by convention and \(\mathbb {R}\) is also closed, as already noted in the example above.

2.

Suppose \(\, x_n \in A\cup B\hspace {0.2cm} \forall n \in \mathbb {N}\, \) and \(\, x_n \longrightarrow x\in \mathbb {R}\). If \(\, x_n \in A\,\) frequently then \(x\in A\) by lemma, since \(A\) is closed. Alternatively, if \(x_n \in B\) frequently then \(x \in B\) again by lemma.
Either way \(x\in A\cup B\). Hence by definition \(A\cup B \) is closed.

3.

Let \(\, S = \bigcap _{i \in I} S_i = \big \{x \in \mathbb {R}:\, x \in S_i \hspace {0.2cm} \forall i \in I\big \}\). Suppose \(\, x_n \longrightarrow x \,\) and \(\, x_n \in S\hspace {0.2cm}\forall n\). Then for each \(\, S_i \in \Omega \, , \, x_n \in S_i\hspace {0.2cm} \forall i \in I\). Thus \(\, x\in S_i \hspace {0.2cm} \forall i \in I\,\), because \(S_i\) is closed \(\, \implies \, x \in S\). Therefore, since \(\, x_n \in S = \bigcap _{i \in I} S_i\hspace {0.2cm} \forall n\, \) and \(\, x\in S, \, S = \bigcap _{i \in I} S_i \) is closed.

Proof. Let \(\, a\leq b\). Then \(\, [a,b]= (-\infty , b]\cap [a,+\infty )\). But \(\, (-\infty , b]\,\) and \(\, [a,+\infty )\,\) are closed sets. Therefore by ‘Theorem (3)’ \([a,b]\) is closed set.

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Proof. Follows by induction on ‘(2) of the previous Theorem’

However, the union of an arbitrary family of closed sets is not closed. For example, if
\(\, A_n = \big [ 1 + 1/n\, , \, 5 - 1/n\big ]\hspace {0.2cm} \forall n \in \mathbb {N}\), then \(\, \bigcup _{n\in \mathbb {N}} A_n = (1,5)\,\) which is not a closed set.

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Proof. \(A = \big \{ a_1\, , \, a_2\, , \, \cdots \cdots , \, a_r\big \} = \big \{a_1\big \} \cup \big \{ a_2\big \}\cup \cdots \cdots \cup \big \{ a_r\big \}\,\) and the singleton sets
\(\, \big \{a_1\big \}\, , \, \big \{a_2\big \}\, ,\cdots \cdots \, \big \{a_r\big \}\,\) are closed. The finite union of closed sets is closed. Hence \(A\) is closed.

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Proof. Let \(x\) be a limit point of \(A\) and let \(x_1\) be a point of \(A\) other than \(x\) contained in \(\, \big ( x - \varepsilon \, , \, x + \varepsilon \big )\). Then, taking \(\, \varepsilon _1 = \min \big \{\big | x_1 - x\big |\, ,\, \varepsilon \big \}\), we have a nbd \(\, \big (x-\varepsilon _1\, , \, x + \varepsilon _1\big ) = I_1\) of \(x\) such that \(\, x_1 \not \in I_1\). By definition, \(I_1\) contains a point of A, say \(x_2\), other than \(x\). Next, taking \(\varepsilon _2 = \min \big \{ \big | x_2 - x\big |\, ,\, \varepsilon _1\big \}\,\) we get a nbd \(\, \big (x - \varepsilon _2\, ,\, x+ \varepsilon _2\big ) = I_2\) of \(x\) such that \(x_1\, , \, x_2, \in I_2\). By definition \(I_2\) contains a point say \(x_3\), of \(A\) other than \(x\). Since \(\, I_3\subset I_2 \subset I_1 \subset \big ( x- \varepsilon \, ,\, x + \varepsilon \big )\), the distinct points \(\, x_1\, , \, x_2\, , \, x_3\in \big (x - \varepsilon \, ,\, x + \varepsilon \big )\). continuing the argument indefinitely, we find many points \(\, x_1\, , \, x_2\, ,\, x_3\, , \cdots \cdots \) of \(A\). Therefore \(\, \big ( x - \varepsilon \, , \, x + \varepsilon \big ) \cap A\,\) is a infinite set.

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Proof. Suppose \(x\) is a limit point of \(A\). Then \(\, \forall , n \in \mathbb {N}\, \exists \, x_n \in A\, \ni \, 0 < \begin {vmatrix} x_n - x\\ \end {vmatrix} < 1/n\). Let \(\varepsilon > 0\) be given. Then \(\, \exists \, m \in \mathbb {N}\, \ni \, M > 1/\varepsilon \). Thus, \(\, \forall \, n > M\, \) we have \[\begin {vmatrix} x_n - x\\ \end {vmatrix} < \frac {1}{n} < \frac {1}{M} < \varepsilon \] Hence, \(\, x_n \longrightarrow x\,\) as \(\, n \longrightarrow \infty \).
Conversely, suppose \(\, \big \{ x_n \big \}^{\infty }_{n = 1}\, \) is a sequence in \(\, A\, \ni \, x_n \longrightarrow x\,\) as \(\, n \longrightarrow \infty \). Let \(\varepsilon > 0\) be given. Then \(\, \exists \, M \in \mathbb {N}\, \ni \, | x_n - x| < \varepsilon \,\, \forall \, n > M\). Therefore, since \(\, x_n \in A\,\) for all \(\, n\in \mathbb {N}\, , \, x\) is a limit point of \(A\).

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Proof.

(i).

We prove that \(\, \overline {A} \supset A\). By definition \(\, \overline {A} = A \cup A' \, \implies \, A \subset \overline {A}\).

(ii).

We prove that if there is another closed set \(\, B \, \ni \, A \subset B\), then \(\, \overline {A} \subset B\).
Suppose \(B\) is closed with \(\, A\subset B\). Let \(\, x \in \overline {A} \, \ni \, x_n \longrightarrow x\,\) where \(\, x_n \in A\,\, \forall n\). Then \(\, x_n \in B\) ( since \(\, A\subset B\)). Therefore \(\, \overline {A} \subset B\).

(iii).

We prove that \(A\) is closed \(\iff \, \overline { A} = A\). Let \(A\) be a closed set. Also, let \(\, x\in \overline {A}\,\) with \(\, x_n \in A\, \, \forall \, n\,\) and \(\, x_n\longrightarrow x\). Since \(A\) is closed, \(\, x\in A\, \implies \, \overline {A} \subset A\). From (i) \(\, A\subset \overline {A} \, \implies , \overline {A} = A\).
Conversely, suppose \(\, \overline {A} = A\). We show that \(A\) is closed. For \(\, x_n \in A\, \forall n\,\) with \(\, x_n \longrightarrow x\,\) in \(\overline {A}\). Then \(\, x\in A\). Therefore \(A\) is closed.

Note. \(A\) is closed if it contains all limit points.

(iv).

We prove that a real number \(\, x\in \overline {A} \, \iff \, \forall \, \varepsilon > 0 \, \exists \, y \in A\, \ni \, \begin {vmatrix} x - y\\ \end {vmatrix} < \varepsilon \).
Suppose \(\, x \in \overline {A}\), where \(\, x_n \longrightarrow x\,\) with \(\, x_n \in A\, \forall \, n\). Then, given any \(\, \varepsilon > 0\, \exists \, m \in \mathbb {N}\, \ni \, \begin {vmatrix} y - x\\ \end {vmatrix} < \varepsilon \).
Conversely, for each positive number \(n\), let \(\varepsilon = 1/n\) and chose \(\, x_n \in A\, \ni \\ \begin {vmatrix} x - x_n\\ \end {vmatrix} < 1/n \, \implies \, x_n \longrightarrow x\,\) as \(n \longrightarrow \infty \). Therefore \(\, x \in \overline {A}\).

(v).

We prove that \(\overline {A}\) is a closed set
We assume that \(\, \forall \, n \, \, x_n \in \overline {A}\,\) and \(\, x_n\longrightarrow x\). We show that \(\, x\in A\). If \(\varepsilon > 0\) is given then \(\, \exists , m\in \mathbb {N}\, \ni \, \begin {vmatrix} x_n - x\\ \end {vmatrix} < \varepsilon /2\,\, \forall \, n > m\). For this \(m\) , chose \(y \in A\) so that \(\, \begin {vmatrix} x_n - y\\ \end {vmatrix} < \varepsilon /2\). Then by triangular inequality, \[\begin {vmatrix} x - y\\ \end {vmatrix} \leq \begin {vmatrix} x - x_n\\ \end {vmatrix} + \begin {vmatrix} x_n - y\\ \end {vmatrix} < \dfrac {\varepsilon }{2} + \dfrac {\varepsilon }{2} = \varepsilon \] Hence, by (iv) above, \(\, x\in \overline {A}\). Therefore \(\overline {A}\) is closed.

Proof. Since \(S\) is bounded and infinite \(\, \exists \,[h,k]\, \ni \, S\subset [h,k]\) and \([h,k]\cap S\,\) is an infinite set.
Let \(\, T = \big \{x: \big (-\infty , x\big ) \cap S\,\) is finite \(\big \}\). Clearly \(\, h \in T \neq \emptyset \,\) and \(K\) is an upper bound of \(T\). For if \(\, x_0\in T\, \ni \, x_0 > K \, \implies \, K\, \) is not an upper bound of \(T\). Hence, by completeness axiom \(T\) has the supremum. Let \(\xi = \sup T\). If \(\, \xi \in (a,b), \, a< \xi \, \implies \, \) a is not an upper bound of \(T\) and so \(\, \exists \, \eta \in T\, \ni \, a< \eta \leq \xi \, \implies \, \big (-\infty , a \big ] \cap S\,\) is finite, for \(\, \big (-\infty , h\big )\cap S \,\) is finite. But \(\, \xi < b \, \implies \, b \not \in T\, \implies \, \big (-\infty \, ,\, b\big ] \cap S\,\) is infinite.
Thus, \(\, (a,b)\cap S - \big (-\infty \, ,\, a\big ] \cap S\,\) is an infinite set. Thus, \(\xi \) is limit point of \(S\).


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