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Can we claim the dimension of the algebraic complement of a Banach space $Y$ in the Banach space $X$ never equals $\aleph_0?$ Here, $Y$ is a linear subspace of $X?$

Arman
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If you have a closed subspace $Y$ of Banach space $X$ ($Y$ will be closed if and only if it is a Banach space in its own right), we can form the quotient space $$X / Y = \{x + Y : x \in X\}.$$ We can equip this space with a norm that will make it a Banach space as well.

Suppose $Z$ is a (not necessarily closed) complement to $Y$ in $X$ (i.e. $X = Y \oplus Z$). We can naturally identify $x + Y$ with $z$, where $y \in Y$ and $z \in Z$ are the unique vectors where $x = y + z$. This map is well-defined, and forms an isomorphism between $X/Y$ and $Z$ as pure (un-normed) vector spaces.

So, if $Z$ has alebraic dimension $\aleph_0$, then so must the Banach space $X / Y$, which is impossible.

Theo Bendit
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  • Thanks, but I mention $Y$ is a linear subspace of $X.$ In fact in my question the norm of Y is not necassarily equals the norm induced from X. – Arman Feb 10 '21 at 05:41
  • Oh, I wish you had made that term more clear. In that case, basically anything goes. If you take an infinite-dimensional Banach space $X$, then it has an uncountable Hamel basis $B$. Remove $\aleph_0$ points from $B$ to get $B'$, then $B$ and $B'$ have the same cardinality. Let $Y = \operatorname{span} B'$. Extend a bijection between $B$ and $B'$ to get a (vector space) isomorphism between $Y$ and $X$, and import the norm from $X$ to $Y$ so that this isomorphism is an isometry. Then $Y$ is a Banach space with countable codimension and unrelated norm inside Banach space $X$. – Theo Bendit Feb 10 '21 at 05:49
  • Yes, in this case (the algebraic completion of Y in X is $\aleph0$) is there possible to extend the norm of Y to a complete norm on X? (X and Y are both Banach spaces with Y is (only) a linear subspace of X, also the norms X and Y are not necassarily relevant) – Arman Feb 10 '21 at 07:18
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    I'm not sure I follow your question, but no, you cannot extend the norm of $Y$ to all of $X$ and expect it to be complete, for the reasons in my answer. That would make $Y$ a closed subspace of this new Banach space $X$ (under its new norm), which would contradict $Y$ having countably infinite codimension. – Theo Bendit Feb 10 '21 at 07:41