# Clarifying definition of $\mathbb{R}\mathbb{P}^n$

This might be a very silly question, but we know that $$\mathbb{R}\mathbb{P}^n$$ is defined as the equivalence classes $$[x]$$ ($$x\in\mathbb{R}^{n+1}$$ nonzero) such that $$x \sim y \iff x = \lambda y$$ for any real, nonzero $$\lambda$$.

So, these elements in $$\mathbb{R}\mathbb{P}^n$$ are lines passing through the origin but without the origin in the line.

My question/problem comes with Lee's definition of $$\mathbb{R}\mathbb{P}^n$$ as the collection of $$1$$-dimensional linear subspaces of $$\mathbb{R}^{n+1}$$. But through this definition, it would make sense that the origin would be included in every equivalence class of $$\mathbb{R}\mathbb{P}^n$$ since $$\textbf{0} \in \text{span}(x)$$ for any $$x\in\mathbb{R}^{n+1}$$.

Why do these two definitions not seem equivalent? What am I missing?

• The "origin" is not a one-dimensional linear subspace of $\Bbb R^{n+1}$. – Lord Shark the Unknown May 2 '19 at 6:53
• @LordSharktheUnknown Yes, I understand that we don't take the origin to it's span since it's span is zero dimensional. But, I guess where the main confusion comes from, is that the span of any set with one vector will include the zero vector (origin). Edit: I'm an idiot. Never mind, thank you. – Good Morning Captain May 2 '19 at 6:59

Just as there are many different constructions of $$\Bbb{R}$$, there are many different constructions of $$\Bbb{RP}^n$$. But the underlying construction is not relevant (most of the time). What matters is that they are isomorphic as projective spaces.