Cauchy Regular and continuity

By | April 27, 2017

Definition. Let \left(X,d_{X}\right),\left(Y,d_{Y}\right) be metric spaces and f:X\rightarrow Y a function. We say f is Cauchy regular provided that any Cauchy sequence \left\{ a_{n}\right\} in X, \left\{ f\left(a_{n}\right)\right\} is a Cauchy sequence in Y .

Note that if f is uniformly continuous on X, then it is Cauchy regular. Indeed, let \left\{ x_{n}\right\} be a Cauchy sequence in X and let \varepsilon>0 be given. Since f is uniformly continuous, there exists \delta>0 such that d_{X}\left(x,y\right)<\delta implies d_{Y}\left(f\left(x\right),f\left(y\right)\right)<\varepsilon.

Since \left\{ x_{n}\right\} is Cauchy, there exists N such that n,m\ge N implies d_{X}\left(x_{n},x_{m}\right)<\delta. So d_{Y}\left(f\left(x_{n}\right),f\left(x_{m}\right)\right)<\varepsilon. Hence \left\{ f\left(x_{n}\right)\right\} is Cauchy in Y.

One may ask whether a Cauchy regular function is continuous. It is clearly true.

Proposition. Let \left(X,d_{X}\right),\left(Y,d_{Y}\right) be metric spaces
and f:X\rightarrow Y a Cauchy regular function. Then f is continuous
on X.

Proof. Suppose not. There exists x_{0}\in X such that f is not continuous at x_{0}. Then there exists \varepsilon_{0}>0 such that for any \delta>0, there exists x\in X such that d_{X}\left(x,x_{0}\right)<\delta but d_{Y}\left(f\left(x\right),f\left(x_{0}\right)\right)\ge\varepsilon_{0}.

By taking \delta=\frac{1}{n}, we can choose x_{n} in X such that x_{n}\rightarrow x_{0} in X. Now define y_{n}=\begin{cases} x_{n} & \text{odd}\\ x_{0} & \text{even} \end{cases}. Since x_{n} is convergent, \left\{ y_{n}\right\} is Cauchy. So \left\{ f\left(y_{n}\right)\right\} is Cauchy. But

    \[ d_{Y}\left(f\left(y_{2n}\right),f\left(y_{2n+1}\right)\right)=d_{Y}\left(f\left(y_{2n}\right),f\left(x_{0}\right)\right)\ge\varepsilon_{0}, \]

shows \left\{ f\left(y_{n}\right)\right\} is not Cauchy, a contradiction.

So Cauchy regular is continuous. Here we note that we didn’t used any kind of continuity.

Recall the Heine-Cantor theorem. If X is compact and f:X\rightarrow Y is continuous, then f is uniformly continuous. One may ask what condition on X gurantees a continuous function f is Cauchy regular. In fact, if we impose X is complete, this is enough.

Proposition. Let f:X\rightarrow Y be continuous. If X is complete, then f is Cauchy regular. In this case, continuity and Cauchy regular is

Proof. Let \left\{ x_{n}\right\} be a Cauchy sequence in X. By completeness of X, there is x\in X such that x_{n}\rightarrow x in X.
Then triangluar inequality shows

    \[ d_{Y}\left(f\left(x_{n}\right),f\left(x_{m}\right)\right)\le d_{Y}\left(f\left(x_{n}\right),f\left(x\right)\right)+d_{Y}\left(f\left(x\right),f\left(x_{m}\right)\right). \]

Since f is continuous at x, there exists N such that n,m\ge N implies

    \[ d_{Y}\left(f\left(x_{n}\right),f\left(x\right)\right)<\frac{\varepsilon}{2}\quad\text{and}\quad d_{Y}\left(f\left(x\right),f\left(x_{m}\right)\right)<\frac{\varepsilon}{2}. \]

Hence there exists N such that n,m\ge N implies

    \[ d_{Y}\left(f\left(x_{n}\right),f\left(x_{m}\right)\right)<\varepsilon. \]

So f is Cauchy regular.

Example. Define f:\mathbb{Q}\rightarrow\mathbb{R}

    \[ f\left(x\right)=\begin{cases} 0 & \text{if }x^{2}<2\\ 1 & \text{if }x^{2}\ge2. \end{cases} \]

Then f is continuous \mathbb{Q}. But this function is not Cauchy regular since it cannot be extended to \mathbb{R} as a continuous function. The example does not contradict our previous theorem since \mathbb{Q} is not complete.

There is a continuous function f which is Cauchy regular, but not uniformly continuous. Consider f:[0,\infty)\rightarrow\mathbb{R} defined by f\left(x\right)=x^{2}. Then it is Cauchy regular, but not uniformly continuous.

Definition. A metric space \left(X,d\right) is Cauchy precompact if every sequence admits a Cauchy subsequence.

Proposition. If X is Cauhcy-precompact and f:X\rightarrow Y is Cauchy regular, then f is uniformly continuous.

Proof. Suppose not. Then there exists \varepsilon>0 and sequences \left\{ a_{n}\right\}, \left\{ b_{n}\right\} such that d_{X}\left(a_{n},b_{n}\right)\rightarrow0
but d_{Y}\left(f\left(a_{n}\right),f\left(b_{n}\right)\right)\ge\varepsilon. Since X is Cauchy precompact, there exists a Cauchy subsequence \left\{ a_{n_{k}}\right\} of \left\{ a_{n}\right\}. Again by Cauchy-precompactness, there exists a Cauchy subsequence \left\{ b_{m_{k}}\right\} of \left\{ b_{n_{k}}\right\}. Define \left\{ c_{k}\right\}
by a_{m_{1}}, b_{m_{1}},a_{m_{2}},b_{m_{2}},\dots. One may check it is Cauchy sequence. Then if k is odd, then

    \[ d_{Y}\left(f\left(c_{k+1}\right),f\left(c_{k}\right)\right)=\left|f\left(a_{m_{k^{\prime}}}\right)-f\left(b_{m_{k^{\prime}}}\right)\right|\ge\varepsilon_{0}. \]

But \left\{ f\left(c_{k}\right)\right\} is Cauchy, a contradiction. Hence f is uniformly continuous on X.

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