# If $X/F$ is separable, must $X/F$ isomorphic to $F$?

Suppose $X$ is a Banach space and $F$ is a closed subspace of $X$. Clearly $X/F$ is a Banach space, equipped with quotient norm.

Question: If $X/F$ is separable, must $X/F$ isomorphic to $F$?

The motivation of this question comes from Corollary $3.4$. It seems that I need this fact to conclude that $F$ is linearly complemented in $X$.

Every separable Banach space is isomorphic to a quotient of $\ell_1$. Certainly most of separable Banach spaces are not isomorphic to a subspace of $\ell_1$.

No, in general, this is very false.

A counterexample can already be obtained by taking $X =\Bbb {R}^3$ and $F=\Bbb {R}^2\times \{0\}$.

Then $\dim X/F=1$, but $\dim F=2$, so that they can not be isomorphic.

But since finite dimensional spaces are always separable and complete, your assumptions hold.

• Then may I know how the author obtained conclusion of corollary $3.4$? – Idonknow Dec 5 '15 at 8:04
• @GEdgar: No, I did not. Every finite dimensional space is complete and hence a Banach space. The question does not say that the spaces under consideration should be infinite dimensional. – PhoemueX Dec 5 '15 at 21:29
• My error. For some reason I thought it said $F = \mathbb R^2 \setminus \{0\}$ instead. – GEdgar Dec 6 '15 at 15:06

In direct sum $X = l^1 \oplus l^2$, take subspace $F = 0 \oplus l^2$, which is reflexive, to get quotient $X / F$ isomorphic to $l^1$, which is separable and not reflexive, and therefore not isomorphic to $F$.