[UnSA] Atsuji Space is Complete
Today, we were looking at the abstract of a paper which says"It is well-known that Atsuji Space is a complete space" .... I was riddled with this statement, I found the proof, It was based from 1955 Paper by Norman Levine and William G. Saunders.
Recall that a metric space (X,d) is called Atsuji Space iff every real-valued continuous functions on X are also uniformly continuous on X.
We want to prove that:
If (X,d) is an Atsuji space then every Cauchy sequence (x_n) in X converges.
Equivalently, we want to prove that
If there exists a Cauchy sequence in X which does not converge, then there exists a continuous function f: X \rightarrow \mathbb{R} which is not uniformly continuous.
In proving the above assertion, we need a following Lemma
[Uryohn's Lemma]
Let X be a normal space. If A and B are two disjoint closed subsets of X, then there exists a continuous function f from X to the real line \mathbb{R} such that f(x)=0 for all x\in A and f(x)=1 for all x\in B.
The proof of Uryohn's Lemma is rather long, I remembered that the nearly two pages proof is on James Munkers Book, Topology. Recall also that from elementary topology course: all metric spaces are normal, so Uryohn's Lemma works also for metric space.
It is not hard to prove that: A Cauchy sequence which has at least one convergent subsequence, is a convergent sequence. Therefore a divergent Cauchy sequence would imply that all of its subsequences are also divergent.
Suppose that \{x_i\}_{i=1}^{\infty} is a Cauchy sequence of X which diverges. Consider the set \{x_1,x_2,\cdots,x_{i},\cdots\}, since (x_n) diverges this set has no limit point on X. Define
A:=\{x_{2k} | k\in \mathbb{N}\} \qquad B:=\{x_{2k-1} | k \in \mathbb{N}\}
The element of A will form a subsequence of (x_n) and by our remark on the previous paragraph this subsequence also diverges, thus A also has no limit point. Similarly B also has no limit point. Therefore A and B are disjoint closed subsets of X, thus by Uryohn Lemma since X is normal space, there exists a continuous function f:X\rightarrow \mathbb{R} such that f(x_{2k-1})=0 and f(x_{2k})=1. This function is not uniformly continuous, since there exists \epsilon_0 = \frac{1}{2} and two sequences u_k = x_{2k} and v_k=x_{2k-1} with \lim (u_n - v_n) =0 (since x_n is Cauchy) but f(u_n)-f(v_n)=1> \frac{1}{2}=\epsilon_0.
Recall that a metric space (X,d) is called Atsuji Space iff every real-valued continuous functions on X are also uniformly continuous on X.
We want to prove that:
If (X,d) is an Atsuji space then every Cauchy sequence (x_n) in X converges.
Equivalently, we want to prove that
If there exists a Cauchy sequence in X which does not converge, then there exists a continuous function f: X \rightarrow \mathbb{R} which is not uniformly continuous.
In proving the above assertion, we need a following Lemma
[Uryohn's Lemma]
Let X be a normal space. If A and B are two disjoint closed subsets of X, then there exists a continuous function f from X to the real line \mathbb{R} such that f(x)=0 for all x\in A and f(x)=1 for all x\in B.
The proof of Uryohn's Lemma is rather long, I remembered that the nearly two pages proof is on James Munkers Book, Topology. Recall also that from elementary topology course: all metric spaces are normal, so Uryohn's Lemma works also for metric space.
It is not hard to prove that: A Cauchy sequence which has at least one convergent subsequence, is a convergent sequence. Therefore a divergent Cauchy sequence would imply that all of its subsequences are also divergent.
Suppose that \{x_i\}_{i=1}^{\infty} is a Cauchy sequence of X which diverges. Consider the set \{x_1,x_2,\cdots,x_{i},\cdots\}, since (x_n) diverges this set has no limit point on X. Define
A:=\{x_{2k} | k\in \mathbb{N}\} \qquad B:=\{x_{2k-1} | k \in \mathbb{N}\}
The element of A will form a subsequence of (x_n) and by our remark on the previous paragraph this subsequence also diverges, thus A also has no limit point. Similarly B also has no limit point. Therefore A and B are disjoint closed subsets of X, thus by Uryohn Lemma since X is normal space, there exists a continuous function f:X\rightarrow \mathbb{R} such that f(x_{2k-1})=0 and f(x_{2k})=1. This function is not uniformly continuous, since there exists \epsilon_0 = \frac{1}{2} and two sequences u_k = x_{2k} and v_k=x_{2k-1} with \lim (u_n - v_n) =0 (since x_n is Cauchy) but f(u_n)-f(v_n)=1> \frac{1}{2}=\epsilon_0.
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