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Let $X$ be a completely regular Hausdorff space. Such a space is also known as Tychonoff space, or a $T_{3.5}$-space. Furthermore, let's assume that $X$ is not compact.

Question. Does $X$ admit a continuous function $f: X\to \mathbb{R}$ with unbounded image?

Context. It is known that if $X$ is a non-compact metric space, then $X$ admits an unbounded continuous real-valued function. This was discussed thoroughly in this MSE thread. Note that the same conclusion holds if $X$ is a non-compact normal space (also known as $T_4$-space). This is because the proof using Tietze extension theorem (see the answer in the linked MSE thread by the user Espace' etale) still works. This is the motivation for asking the current question.

In general, Tietze extension theorem fails for $T_{3.5}$-spaces (for example, consider the Moore plane), so one cannot apply the trick above; of course, it is possible that the question has a negative answer, in which case I'd love to see a counter-example.

Prism
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    The linked question does not specifically ask for Tychonoff spaces but several of the examples in the answers are Tychonoff. – Eric Wofsey Jul 10 '20 at 01:11

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Not necessarily. A space $X$ such that every continuous $f:X\to\Bbb R$ is bounded is called pseudocompact. Every countably compact space is pseudocompact, and $\omega_1$ with the order topology is a countably compact, non-compact Tikhonov (and even hereditarily normal) space, so it is also a pseudocompact space that is not compact. (Note that the argument by Espace' etale uses more than normality: it also uses the fact that sequential compactness implies compactness in metric spaces.)

Brian M. Scott
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