2
$\begingroup$

Let $X$ be a complete normed space and assume the normed space $Y$ is isometric to $X$. Show that $Y$ is complete.

I tried:

Since X is complete $||x_n-x_m||<\epsilon, \forall n,m>N$ and since $Y$ is isometric to $X$ there exists an isometry $f:X\to Y$ such that $$||f(x_n)-f(x_m)||=||x_n-x_m||<\epsilon, \forall n,m>N.$$ I stuck at this step.

$\endgroup$
3
$\begingroup$

Pick an arbitrary Cauchy sequence $\{y_n\}\subset Y$, and let $f\colon X\to Y$ be the isometry. For each $n\geq1$, $y_n=f(x_n)$ for some $x_n\in X$. We have \begin{equation*} \|y_n-y_m\|=\|f(x_n)-f(x_m)\|=\|x_n-x_m\|, \end{equation*} so that $\{x_n\}$ is a Cauchy sequence in $X$. Since $X$ is complete, $x_n\to x$ for some $x\in X$. Therefore, $y=f(x)$ is the limit of $\{y_n\}$, since \begin{equation*} \|y_n-y\|=\|f(x_n)-f(x)\|=\|x_n-x\| \end{equation*}

Therefore, $Y$ is also complete.

$\endgroup$
3
$\begingroup$

Ok so let $\{y_n\}$ be a Cauchy sequence in $Y$. Write $y_n=f(x_n)$ for some $x_n\in X$. As $f$ is an isometry, $\{x_n\}$ is a Cauchy sequence in $X$ and thus has a limit $x$, say. Then it is immediate that $f(x)$ is the limit of $\{y_n\}$. (The result more generally holds for metric spaces. I also note that some people don't require isometries to be surjective, here of course you assume that they are.)

$\endgroup$
2
$\begingroup$

There is nothing to prove here: $Y$ is a complete metric space even if it wouldn't be a vector space. It's all in the word "isometric". This word says that $Y$ is a bijective copy of $X$ whereby the distance between points is preserved. This implies that the notions of "convergence" or "Cauchy sequence" in $X$ and in $Y$ are the same.

$\endgroup$
1
$\begingroup$

You've got the idea, but your proof is lacking. What is $\varepsilon$? What are $x_m$ and $x_n$?

Here's my suggestion: fix an isometry $f$ from $Y$ to $X$, let $(y_n)_{n=1}^\infty$ be a Cauchy sequence in $Y$, transport it to $X$ via $f$, and show that this new sequence is Cauchy in $X$. Since $X$ is complete, $(f(y_n))_{n=1}^\infty$ has a limit $x$ in $X$. Since $f$ is surjective, $x = f(y)$ for some $y$ in $Y$. Now show that $y_n \to y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.