Is a countable union of complete subspaces complete?

Question

I would like to ask the following, which I wanted to use a part of my proof but couldn't determine if it's right:

Assume $X$ is a normed space, and $(X_n)_{n\in \mathbb N}$ complete subspaces.

Must $\bigcup_{n\in \mathbb N} X_n$ be complete?

My intuition is that it should be right because at some point, every cauchy sequence should be in $X_j$ for a specific $j$, and therefore converge. I couldn't show that it must be in one of those starting at some point.

Thanks!

Answer 1

It's not true. Let $X$ be the space of sequences that are eventually $0$ with the sup norm. Let $X_n$ be the subspace of $X$ consisting of the sequences $x$ with $x(i)=0$ if $i\ge n$.

Consider the sequence $(x_n)$ with $x_n=(1,1/2,1/3,\ldots,1/n,0,0,\ldots)$.

Answer 2

Easy finite-dimensional counterexample: $X = \mathbb{R}$, and $X_n = [\frac{1}{n},1]$. Then each $X_n$ is complete, and

$$ \bigcup_{n=1}^\infty X_n = (0,1] $$

which is quite clearly not complete.

If you're looking for normed linear subspaces, consider $X = \mathbb{R}^2 = \mathbb{C}$, and $X_n = e^{i\pi/n}\mathbb{R}$, the real axis rotated by $\pi/n$. Then $$ \bigcup_{n=1}^\infty X_n $$ does not contain the real axis, but it is easy to find a Cauchy sequence that should converge to a point on the real axis.