Cauchy sequence and bijective function

Question

If $\{fx_n\}$ is a Cauchy sequence with $f$, a continuous bijective self-mapping on a complete metric space $X$, then $\{x_n\}$ is Cauchy?

I guess it is. Since the metric space is complete and $f$ is bijective, $$\lim_{n\rightarrow\infty}fx_n=fx$$ for some $x$ in $X$.

By the continuity of $f$, $$\lim_{n\rightarrow\infty}fx_n=f\left(\lim_{n\rightarrow\infty}x_n\right)=fx$$ so that $\lim_{n\rightarrow\infty}x_n=x$ and hence $\{x_n\}$ is Cauchy.

Is this reasoning complete? Thanks.

No, your reasoning is circular: you are assuming that $x_n$ converges when you claim that $\lim_{n\to\infty} f(x_n)=f\left(\lim_{n\to\infty} x_n\right)$, which is the same as (actually, stronger than) saying that $x_n$ is Cauchy. — , May 02 '20 at 12:06
what about continuity being replaced with uniform continuity? — mark haokip, May 02 '20 at 12:20
I meant if $f$ is uniformly continuous and bijective, then ${x_n}$ is Cauchy? — mark haokip, May 02 '20 at 12:30

score 1 · Accepted Answer · 2020-05-02T14:27:42.940

Your argument is circular and the claim is false.

In the same spirit as this answer, which is only missing the detail of $X$ being complete, call $R_n=\{n\}\times[0,\infty)$, $C_n=\{(x,y)\in\Bbb R^2\,:\, (x-n)^2+(y-1)^2=1\}$, $L=\Bbb R\times\{0\}$ and consider the metric space $X\subseteq \Bbb R^2$ $$X=L\cup\left(\bigcup_{n\ge 0} R_{-4n}\right)\cup\left(\bigcup_{n\ge 1} C_{4n}\right)$$

Then consider this map $$f(x,y):\begin{cases} (x+4,y)&\text{if }(x,y)\notin R_{0}\\ \left(x+4+\sin\left(4\arctan y\right), 1-\cos(4\arctan y)\right)&\text{if }(x,y)\in R_0\end{cases}$$

In other words, $f$ translates everything on the right by four, except the last closed ray $R_0$, which is winded counterclockwise around the first circle $C_4$. You can check $X$ is closed in $\Bbb R^2$, and thus complete, and that the function $f:X\to X$ is uniformly continuous and bijective, but $f^{-1}$ is not continuous.

Cauchy sequence and bijective function

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