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In part 6) of example 103 in Steen & Seebach's "Counterexamples in Topology" they outline a proof that uncountable products of $\mathbb{Z}$ ($X_{\lambda}=\prod_{\alpha\in A}Z^+_{\alpha}$) are not normal by trying to show that any neighbourhoods U & V of $P_0=\{x:\text{only 0 may appear more than once among the coordinates of x}\}$ & $P_1=\{x:\text{only 1 may appear more than once among the coordinates of x}\}$ must intersect. They define a sequence of finite index sets $F_n\subset A$ and a sequence of points $x^n\in X_{\lambda}$, and define $F_n(x^n)=\cap_{\alpha\in F_n}\pi_{\alpha}^{-1}(x^n)$ (i.e. the set of all points in $X_{\lambda}$ that are the same as $x^n$ on the indices in $F_n$). Then they say that $F_n(x^n)\subset U$. WHY?

Surely $F_n(x^n)$ may contain points with any non-zero integer in indices outside of $F_n$, and those numbers may repeat which would exclude the point from $U$. What am I misunderstanding here?

Ben
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    The title and the body are in conflict. Are we talking about $\mathbb{R}$ or about $\mathbb{Z}$? – saulspatz Jun 13 '21 at 15:51
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    It's likely that $U$ is defined as an open set (probably including one of the $P_i$). If so, then the definition of the product topology will produce some $F_n$'s and some $x_n$'s with the claimed inclusions $F_n(x^n)\subset U$. A more detailed answer to the question "Why?" would require knowing what $U,x^n,F_n$ are, rather than just going with my guesses. – Andreas Blass Jun 13 '21 at 18:12

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An answer is given by Brian M. Scott's comments at the bottom of this answer: https://math.stackexchange.com/a/65458/190762

Ben
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