TITLE: Is the space R^N convex?
QUESTION [5 upvotes]: My question is if $\mathbb{R}^{N}$ is convex.
The definition I have for a set S to be convex is that if any convex combination of any two elements of S is in S, where a convex combination is defined as follows:  
Given vectors $x,y\in \mathbb{R}^{N}$, a vector $z\in\mathbb{R}^{N}$ is a convex combination of x an y if there exists a scalar $\alpha \in [0,1]$ such that $z=\alpha x + (1-\alpha)y$.
Intuitively, I'd say that the answer is yes, my reasoning being that scalar multiplication and addition do not change the dimension of a vector, so if $x,y\in \mathbb{R}^{N}$ then $\alpha x + (1-\alpha)y$ must also be in $\mathbb{R}^{N}$, so all convex combinations of any two elements in $\mathbb{R}^{N}$ must also be in $\mathbb{R}^{N}$ and thus $\mathbb{R}^{N}$ is convex, though I am not sure that this would be sufficient to prove the concept.  Any tips?

REPLY [4 votes]: What you have is basically fine. If you want to make it a bit more rigorous, let $x=\langle x_k:k\in\Bbb N\rangle$ and $y=\langle y_k:k\in\Bbb N\rangle$ be arbitrary elements of $\Bbb R^{\Bbb N}$, and let $\alpha\in[0,1]$ be arbitrary. Then
$$\begin{align*}
\alpha x+(1-\alpha)y&=\langle\alpha x_k:k\in\Bbb N\rangle+\langle(1-\alpha)y_k:k\in\Bbb N\rangle\\
&=\langle\alpha x_k+(1-\alpha)y_k:k\in\Bbb N\rangle\;,
\end{align*}$$
which is clearly in $\Bbb R^{\Bbb N}$.