It is well-known that if $X$ is a Banach space and $Y$ is a closed subspace then $X /Y$ with quotient norm $$ \| [x] \|_{X/Y} = \inf_{y \in Y} \| x-y \| $$ is a Banach space.

I am trying to prove the converse:

Let $X$ be a normed space, $Y$ its closed subspace which is complete. If $X/Y$ is complete spce, then $X$ is complete.

My attempts was to equivalently prove that $X$ is not complete and conclude that $X/Y$ cannot be complete. Assume that there exists sequence $(x_n)$ in $X$ which is Cauchy, but not convergent. Using the inequality $$ \| [x_n ] \|_{X/Y} \leq \| x \|_X $$ we obtain that sequence $ ([x_n])$ is Cauchy in $X/Y$. Next step, that I cant make would be to prove that it is not convergent, or that this allows us to construct non-convergent Cauchy sequence.

I also proved that $Y$ has to be complete, I have a counterexample otherwise. So this fact has to be used somehow.

Note that this is an exercise for first weeks of introductory functional analysis course, so I can't use any advanced techniques.


Since $(x_n)$ is Cauchy then $([x_n])$ is Cauchy. Therefore it converges, say $[x_n]\to [z]$ in $X/Y$.

By the definition of the norm, for every $m\in\mathbb{N}$ you can find $n_m\in\mathbb{N}$ and $y_m\in Y$ such that $$\left\|{x_{n_m}-z-y_m}\right\|<\frac{1}{m}.$$

Then prove $(y_m)$ is Cauchy, so it converges, and also does $(x_{n_m})$.

  • $\begingroup$ @wroobell How did you prove that $(y_m)$ is Cauchy? $\endgroup$ – richarddedekind May 25 '17 at 3:07
  • $\begingroup$ Yes why is that sequence cauchy? I'm also stuck here. Please help. $\endgroup$ – MathCosmo May 10 '19 at 9:38

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