Show that $f$ can be extended to a Cauchy-sequence preserving continuous mapping on $\overline{A}$.

Question

Question: Let $(X,d)$ be a metric space and $A\subset X.$ If $f: A\to\mathbb R$ be a Cauchy-sequence preserving continuous mapping then show that $f$ can be extended to a Cauchy-sequence preserving continuous mapping on $\overline{A}$.

I have tried it as follows: Let $x\in\overline{A}.$ Then there exists $(x_n)\subset A$ such that $\lim x_n=x.$ Thus $\lim f(x_n)$ exists, $(f(x_n))$ being Cauchy.

Let $f_*:\overline{A}\to\mathbb R$ be such that $f_*(a)=\lim f(x_n)$ where $(x_n)\subset A$ with $\lim x_n=a.$

I have proved that $f_*$ is well-defined.

However I could not show $f_*$ is Cauchy-sequence preserving continuous mapping.

Please help me.

Answer 1

Some extensive hints:

• Let $(\bar x_n)\subset \bar A$ be a Cauchy sequence.

• There exists a sequence $(x_n)\subset A$ such that $d(f(x_n), f_*(\bar x_n)) \to 0$ and $d(x_n, \bar x_n) \to 0$ (why?).

• From $d(x_n, \bar x_n)\to 0$ it follows $(x_n)$ is Cauchy (why?).

• As $(x_n)$ is Cauchy, $(f(x_n))$ is Cauchy by assumption. Therefore $(f_*(\bar x_n))$ is Cauchy (why again?)

Answer 2

Firstly, you missed something important here: you have a good candidate to be $f_*(x)$ but you did not show that for any converging sequence $x_n \to x$, the limit of $f(x_n)$ is the same. Thus, $f_*(x)$ might be not well defined! However, this can be fixed easily.

Then, you can prove your claim, that is $f_*$ is cauchy preserving. For that, take any Cauchy sequence in $\overline{A}$. Prove that there exists a Cauchy sequence in $A$ that is close to it. Then, the result follows from straightforward calculations.

Edit I have not seen at first that you wrote "I have proved that $f_*$ is well defined. Sorry!

Answer 3

Let us first show $f_*$ is continuous. So let $(x_n)$ be a sequence in $\overline{A}$ such that $x_n \to x \in \overline{A}.$ We aim to show $f_*(x_n) \to f_*(x).$ For each $n,$ there exists a sequence $(x_n^k)_k$ in $A$ such that $x_n^k\to x_n$ as $k \to \infty.$ Thus $$f_*(x_n)=\lim_k f(x_n^k)\tag{1}.$$

Since $(x_n^k)\to x_n,$ there exists $k_n\in \mathbb{N}$ such that $$d(x_n^{k_n},x_n)<\frac{1}{n}.$$ Let $y_n=x_n^{k_n} \in A.$ Then $$d(y_n,x)\leq d(y_n,x_n)+d(x_n,x)<\frac1n+d(x_n,x)\to 0 \text{ as } n \to \infty.$$ As a result $f(y_n) \to f_*(x)$ as $n \to \infty.$

Using $(1),$ there exists $j_n\in \mathbb{N}$ such that $$|f(x_n^{j_n})-f_*(x_n)|<\frac1n \text{ and }|f(x_n^{j_n})-f(y_n)|<\frac1n$$ since $(f(x_n^k))_k$ is Cauchy.

Finally, a triangle inequality argument implies $f_*(x_n)\to f_*(x)$ as $n\to \infty$ and so $f_*$ is continuous.

Next, we show $f_*$ is Cauchy-preserving. Let $(x_n)$ be a Cauchy sequence in $\overline{A}.$ Then as before, for each $n,$ there exists a sequence $(x_n^k)_k$ in $A$ such that $x_n^k\to x_n$ as $k \to \infty$ and $$f_*(x_n)=\lim_k f(x_n^k).$$ Now choose $y_n$ as above and note that $(y_n)_n$ is Cauchy, hence $(f(y_n))_n$ is Cauchy. Finally $$|f_*(x_n)-f_*(x_m)|\leq |f_*(x_n)-f(y_n)|\leq |f(y_n)-f(y_m)|+|f(y_m)-f_*(x_m)|\to 0$$ as $n,m \to \infty.$