Does completeness of $X/Y$ imply completeness of $X$ ? Where $X$ is normed space and $Y$ closed inifinty dimensional subspace of $X$. I tried take $c_{00}$ (space of the finite sequence with supremum norm) and for the subspace every sequence that have $0$ on the first coordinate but I can't prove that this quotient space is complete.


Indeed, you can construct a counterexample with $c_{00}$.

Equip $c_{00}$ with any $\|\cdot\|_p$ norm and let $Y = \{(x_n)_n \in c_{00} : x_1 = 0\}$. Then $Y$ is the kernel of the bounded linear functional $f : c_{00} \to \mathbb{C}$, given by $f(x_n)_n = x_1$. Hence $Y$ is closed in $c_{00}$ and the induced map $\tilde{f} : c_{00}/Y \to \mathbb{C}$ given by $\tilde{f}((x_n)_n + Y) = x_1$ is an isometric isomorphism.

So $c_{00}/Y \cong \mathbb{C}$, which is finite-dimensional and hence Banach.

The statement is true if $Y$ itself assumed to be complete.

Let $(x_n)_n$ be a Cauchy sequence in $X$. Then

$$\|(x_m+Y) - (x_n+Y)\| = \|(x_m - x_n) + Y\| \le \|x_m-x_n\|$$

so $(x_n + M)_n$ is a Cauchy sequence in $X/Y$ so it converges to an element $x + M$.

We have $\|(x_n -x) + M\| = \|(x_n+M) - (x + M)\| \to 0$ so for every $n \in \mathbb{N}$ there exists $p(n) \in \mathbb{N}$ and $y_n \in Y$ such that

$$\|x_{p(n)} + y_n - x\| < \frac1n$$

Notice that $(y_n)_n$ is Cauchy in $Y$

$$\|y_m-y_n\| \le \|y_m - x_{p(m)} - x\| + \|x-x_{p(n)} - y_n\| + \|x_{p(m)} - x_{p(n)}\| \xrightarrow{m,n\to\infty} 0$$

so it converges to an element $y \in Y$.

$$\|x_{p(m)} + y - x\| \le \|x_{p(m)} + y_n - x\| + \|y - y_n\| \xrightarrow{n\to\infty} 0$$

so $(x_{p(n)})_n$ converges to $x-y$. Since $(x_n)_n$ is Cauchy, it also converges to $x-y$.


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