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I have seen from other questions on this site, such as this one, that a regular space would be sufficient, however we haven't yet learned about Lindelöf spaces in the course, so if I want to use the fact that regular Lindelöf spaces are normal, I would have to prove it myself. Surely there is a proof with a Tychonoff space rather than just a regular space, which doesn't use this fact. Maybe there is an easier way of proving such a space is normal?

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  • $\begingroup$ my writeup here gives a lemma as first statement that immediately implies that a countable and regular space is normal. $\endgroup$ – Henno Brandsma Feb 13 '18 at 8:57
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You don't need normality to get uncountability of $X$, if you have Tychonoff.

If $X$ is connected and Tychonoff, then take point $p \in X$ and some non-trivial open set $O$ that contains $p$ (so $X \neq O$). Then being Tychonoff gives us a continuous function $f: X \to [0,1]$ (or to $\mathbb{R}$) such that $f(p) = 0$ and $f[X\setminus O] = \{1\}$, or the other way round, the only important fact is that $0,1 \in f[X]$. From the last fact it follows by connectedness of $f[X]$ that $[0,1] \subseteq f[X]$ because if $C$ is connected in the reals or $[0,1]$ it must be order convex ($c_1 < x <c_2 , c_1,c_2 \in C$ implies $x \in C$). So $f[X]$ is uncountable and hence so is $X$.

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