3.3 Closed Subspaces and the Nested Set Theorem
Two consequences of completeness, both used in the fixed point arguments later in this section. The first says completeness is inherited by closed subsets; the second is the metric-space form of the nested interval property of \(\mathbb {R}\).
Theorem 3.20. If \((X,d)\) is a complete metric space and \(A\) is a closed subset of \(X\), then \((A,d)\) is also complete.
Proof. Let \((x_n)\) be a Cauchy sequence in \(A\subset X\). Then \((x_n)\) is also a Cauchy sequence in \(X\). But \(X\) is complete, so \((x_n)\) converges to say \(x\in X\). Hence \(x\) is a limit point of \(A\). By hypothesis, \(A\) is closed i.e it contains all its limit points. Thus \(x\in A\). Hence we have that \((A,d)\) is complete. □
Theorem 3.21. Let \((X,d)\) be a complete metric space. Let \(\{F_n\}\) be a sequence of nested closed sets in \(X\) with \(\diam (F_n)\rightarrow 0\) as \(n\rightarrow \infty \). Then \(\displaystyle {\bigcap ^{\infty }_{i=1}F_i} \hspace {0.2cm}\) is a singleton element.
Proof. Suppose \(\{F_n\}\) is a sequence of nested intervals each of which are non-empty such that \(\diam \{F_n\}>0\) as \(n\rightarrow \infty \).
Let \(x_n\in F_n\), if \(m>n\), \(F_n\supset F_m\) also \(x_n,x_m\in F_n\). Using the definition of diameter of a set, we have \(d(x_m,x_n)\leq \diam \{F_n\}\).
Now, since \(\diam \{F_n\}\rightarrow 0\) as \(n\rightarrow \infty \), it follows that \(d(x_m,x_n)\rightarrow 0\) as \(m,n\rightarrow \infty \).
Thus \((x_n)\) is a Cauchy sequence in \(X\). Since \(X\) is complete, \(x_n\rightarrow x\in X\).
\(\{x_m,x_{m+1},x_{m+2},\dots \}\subset F_m\) and this sequence converges to \(x\). Hence \(x\in F_n\) for every \(n\). i.e \(\displaystyle {\bigcap ^{\infty }_{n=1}F_n}\hspace {0.2cm}\) is a singleton \(x\). □
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