[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 $f(u_n)-f(v_n)=1> \frac{1}{2}=\epsilon_0$.
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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 $f(u_n)-f(v_n)=1> \frac{1}{2}=\epsilon_0$.