Proposition 6

The space $P_{n,\ p}$ is Hausdorff.

We show first that the relation $\Delta_{n,p}$ is open. If $U$ is an open set in $L_{n+1,p+1}$, to saturate $U$ with respect to $\Delta_{n,p}$ we have to take the union of the images of $U$ under the mapping $X \to T.X$, where $T$ runs through the set of non-singular square matrices of order $p+1$; since each of these mappings is bicontinuous, all these images are open sets and therefore so is their union.

By Proposition 8 of Chapter I, § 8, no. 3, the proof will be complete if we show that the subset $N$ of $L_{n+1,p+1} \times L_{n+1,p+1}$, defined by $\Delta_{n,p}$, is closed. Let $(X, Y)$ be a point of the product space which lies in the closure of $N$, and let $\sigma$ be a sequence of indices such that $X_\sigma$ is non-singular: since $A_\sigma$ is open, there is a neighbourhood $V$ of $(X, Y)$ such that, for each pair $(X', Y') \in N \cap V$, the matrix $X'_\sigma$ is non-singular; as $(X', Y')$ tends to $(X, Y)$ while remaining in $N$, the matrix $Y'_\sigma {X'_\sigma}^{-1}$ therefore tends to $T = Y_\sigma X_\sigma^{-1}$; since we have $Y' = (Y'_\sigma {X'_\sigma}^{-1}) X'$, we see by passing to the limit that $Y = T.X$, and the proof is complete.