Algebraic complements in $\Bbb R^2$

Question

Definition: $Z$ is called an algebraic complement of $Y$ in $X$ if $X=Y\oplus Z$.

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Example: $Y=\Bbb R$ is a subspace of the Euclidean plane $\Bbb R^2$. "Clearly, $Y$ has infinitely many algebraic complements in $\Bbb R^2$."

I don't find this entirely clear, unless this is it:

$Y = \Bbb R = \{(y,0): y\in \Bbb R\}$, $Z_a=\{(a,z):z\in \Bbb R\}$, then we obtain $X=Y\oplus Z_a, \forall a\in \Bbb R$, is that what they are getting at?

Answer 1

As Najib says, the set $Z_a$ is not a linear subspace whenever $a\neq0$.

However, yes, the space $Y$ in your question has infinitely many algebraic complements in $\mathbb{R}^2$. In fact, for any $v\in\mathbb{R}^2\setminus Y$, we have $\mathbb{R}^2=Y\oplus\mathrm{span}(v)$.