$X$ is complete and $T$ is an onto open map, then $Y$ is complete

Question

I am trying to prove sort of a converse to open mapping theorem. If $X, Y$ are normed linear spaces where $X$ is complete, and $T \in B(X, Y)$ is open, onto then I have to show Y is complete. I came across this answer here: https://math.stackexchange.com/a/1446979/698050 and my doubt is the following:

I have proved the proposition that is mentioned, it says $∀y ∈ Y$ there is an $x \in T^{−1}(y)$ such that $∥x∥_X ≤ C ∥y∥_Y$.

But now to prove pullback of a Cauchy sequence in $Y$ is Cauchy in $X$, say I have $x_n$ and $x_m$ in $X$ having their norms bounded by corresponding $C||y_n||$ and $C||y_m||$. This does not imply $||x_n - x_m|| \le C||y_n - y_m||$ as we can have some other $x \in X$ satisfying the proposition for $(y_n - y_m)$. How do I conclude the sequence in $X$ is Cauchy then?

Answer 1

Hint for a different approach: Define $S:X|Ker (T) \to Y$ by $S(x+Ker(T))=Tx$. Verify that $S$ is also an open map. It is also an injective continuous linear map, so it is an isomorphism. Since $X|M$ is complete for any close subspace $M$ of $X$ it follows that $X|Ker (T)$ is complete. Since an isomorphic image of any complete space is complete we are done.

Answer 2

You need to do a little bit more. Since $(y_n)$ is a Cauchy sequence in $Y$. Then there is a subsequence $y_{k_n}$ of $(y_n)$ such that $$\tag{1} ||y_{k_{n+1}}-y_{k_n}||<\frac{1}{2^n}$$

for all $n$. Now, take $x_n\in X$ such that $Tx_n=y_{k_{n+1}}-y_{k_n}$ and $||x_n||\leq C||y_{k_{n+1}}-y_{k_n}||$. Then by $(1)$ it follows that $\sum_{n=1}^{\infty}||x_n||<\infty$. By the completeness of $X$ there is some $x\in X$ such that $s_n\to x$ where $s_n$ are the partial sums of $\sum_{n=1}^{\infty}x_n$. Then for all n $$Ts_n=y_{k_{n+1}}-y_{k_1}$$ and by the continuity of $T$, $Ts_n\to Tx$. It follows that $y_{k_{n+1}}\to Tx+y_{k_1}$, which means that $(y_n)$ has a convergent subsequence, hence $(y_n)$ converges. So, $Y$ is complete.