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Let $(X,\leq)$ be a well-ordered set that contains exactly one element $x$ such that $y<x$ for uncountably many values of $y\in X$. Let $f(x)=1$ and $f(y)=0$ for all other values of $y\in X$. For the interval topology $\tau$ on $X$, show that for every sequence $u_n\to u$ in $X$, $f(u_n)\to f(u)$.

This is what I've tried. Let $u=x$ and $u_n\to u$. Let $B$ be a neighborhood of $f(u)=1$. It is enough to consider $B=\{1\}$. We need to show that there exists an $m$ such that $u_n=x$ for all $n\geq m$. Suppose there exists a subsequence $(u_{n_k})$ such that $u_{n_k}<x$ for all $k$. Then for each $k$ there exist uncountably many $w_k$ such that $u_{n_k}< w_k<x$, but I don't know how to proceed from here.

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Try the other way. $[0,u_{n_k})$ is countable for all $k$ ($0$ is the smallest element of $X$, you could call it $Jim$ if you don't like $0$). So $[0,\sup u_{n_k})$ is? – Daniel Fischer Nov 7 '13 at 18:33
up vote 1 down vote accepted

HINT: Suppose that $\langle y_n:n\in\Bbb N\rangle$ is a sequence such that $y_n<x$ for all $n\in\Bbb N$. For each $n\in\Bbb N$ let $A_n=\{y\in X:y<y_n\}$. Each $A_n$ is countable, so $A=\bigcup_{n\in\Bbb N}A_n$ is countable, and $X\setminus A\ne\varnothing$. Let $y=\min(X\setminus A)$. Show that $y<x$, and use this to show that $\langle y_n:n\in\Bbb N\rangle$ does not converge to $x$.

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Thank you, $y$ is the element I was looking for. – nokiddn Nov 7 '13 at 18:42
@nokiddn: You’re welcome. – Brian M. Scott Nov 7 '13 at 18:43

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