# Is The Space of Probability Vectors Closed With Respect to the $L^1$ norm?

We know that the sequence of probability vectors $x_0, x_1, \ldots$ satisfying $x_{k+1} = P x_k$, where $P$ is a stochastic matrix will always converge to an eigenvector of eigenvalue 1 as $k \rightarrow \infty$. Is it true that the limit of any convergent sequence of probability vectors will also be a probability vector? In general is it true that $\lim_{n \rightarrow \infty} \|x_n\| = \left\| \lim_{n \rightarrow \infty} x_n \right\|$ in a normed space if this is true?

I'm having trouble thinking about this as I can't use Cauchy-Schwarz in a normed space, and didn't learn algebra as well as I should have.

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Bah sorry stupid question we have $\|x_n\| \leq \|x_n-x\| + \|x\|$ and $\|x_n\| \geq \|x\| - \|x_n -x\|$ by the triangle inequality, so the second statement is true, and it implies the first one. Sorry for wasting your time.

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Your vectors $x_k$ presumably lie in a finite dimensional vector space (since $P$ is a matrix), though the dimension is not explicit. In this case the $L^1$ norm is the most natural one to use with "probability vectors", nonnegative vectors whose components sum to 1. It makes that a closed set, but since all norms on a finite dimensional vector space give equivalent topologies, it could be any norm. – hardmath Oct 16 '12 at 14:06