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?


1 Answer 1


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.

  • $\begingroup$ Thanks for the clarification. But why is it called the complement of $Y$? I find that confusing. $\endgroup$
    – JustANoob
    Commented Oct 19, 2021 at 13:56

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