# Every finite-dimensional subspace of $Y$ of a normed linear space $X$ has a closed complement.

The following is Corollary 4' in the book Functional Analysis, Peter D. Lax, chapter 8.

Every finite-dimensional subspace of $$Y$$ of a normed linear space $$X$$ has a closed complement.

I'm struggling to see where my counter-example is wrong.

Let $$X$$ be $$\mathbb{R²}$$ with the euclidean norm and $$Y = \{ (x,x) | x \in \mathbb{R}\}$$. Then, according to above corollary, the set $$Y^c = \{ (x,y) \in \mathbb{R²}| x \ne y\}$$ should be closed, right?

But it is not, since for example the sequence $$x_n := (1, 1- \frac{1}{n}) \in Y^c$$ does not converge in $$Y^c$$.

What am I missing?

There is a misunderstanding ! You have to show: if $$\dim Y < \infty$$, then there is a subspace $$Z$$ of $$X$$ such that $$X= Y \oplus Z$$ and $$Z$$ is closed.
• Thanks for the clarification. But why is it called the complement of $Y$? I find that confusing. Oct 19, 2021 at 13:56