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ADVANCED MATH Prove the following: Theorem. If J is uncountable, then $\mathbb{R}^{J}$ is not normal. Proof (This proof is due to A. H. Stone, as adapted in [S-S.) Let $x=\left(\mathbb{Z}_{+}\right)^{J}$; it will suffice to show that X is not normal, since X is a closed subspace of $\mathbb{R}^{J}$. We use functional notation for the elements of X, so that the typical element of X is a function $\mathbf{x}: J \rightarrow \mathbb{Z}_{+}$ (a) If $\mathbf{x} \in X$ and if B is a finite subset of J, let U(x, B) denote the set consisting of all those elements y of X such that $\mathbf{y}(\boldsymbol{\alpha})=\mathbf{x}(\boldsymbol{\alpha})$ for $\boldsymbol{\alpha} \in \boldsymbol{B}$. Show the sets U(x, B) are a basis for X. (b) Define $P_n$ to be the subset of X consisting of those x such that on the set $J-x^{-1}(n)$ the map x is injective. Show that $P_1$ and $P_2$ are closed and disjoint. (c) Suppose U and V are open sets containing $P_1$ and $P_2$, respectively. Given a sequence $\alpha_{1}, \alpha_{2}, \dots$ distinct elements of J, and a sequence $0=n_{0}<n_{1}<n_{2}<\cdots$ of integers, for each $i \geq 1$ let us set $B_{i}=\left\{\alpha_{1}, \cdots, \alpha_{n_{1}}\right\}$ and define $\mathbf{x}_{i} \in X$ y the equations $\mathbf{x}_{i}\left(\alpha_{j}\right)=j$ for $1 \leq j \leq n_{i-1}$, $\mathbf{x}_{i}(\alpha)=1$ for all other values of $\alpha$. Show that one can choose the sequences and so that for each i, one has the inclusion $U\left(\mathbf{x}_{i}, B_{i}\right) \subset U$. Verified answer
ADVANCED MATH Verified answer
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