Definition: A topological space $X$ which is $T_1$ space is called Tychonoff space or completely regular space ($T_{3\frac{1}{2}}$ space) if for any point $x\in X$ and $A$ closed in $X$ with $x\notin A$ exists continuous function $f:X\to [0,1]$ such that $f(x)=1$ and $f(A)=\{0\}$.

Let me ask you a stupid question please: If we want to show that some topological space $X$ is Tychonoff space we always have to consider closed set $A$ in $X$ which is not empty right?

Because if $A=\varnothing$ then $f(\varnothing)=\{0\}$ does not make sense, right?

  • $\begingroup$ The whole point of Tychonoff as a property is to ensure that we have "enough" continuous functions on $X$, that we can "control" in this way. If we allow $A=\emptyset$ we could take $f \equiv 1$ and we've gained no new info on continuous function whatsoever. (assuming we use $f[A]\subseteq \{0\}$ instead of equal). So it's safe to assume $A$ is non-empty; we lose nothing. Wikipedia requires $f\restriction A\equiv 0$ which is also OK for the empty set ($\forall x \in A: f(x)=0$).. $\endgroup$ – Henno Brandsma Oct 26 '19 at 21:41


Though you could circumvent this problem if you require $f(A)\subseteq\{0\}$, instead of $f(A)=\{0\}$. If $A$ is empty then $f(\varnothing)=\varnothing\subseteq\{0\}$.

The definition requires that your could find a suitable $f$ (in general depending on $A$ and $x$) for every $x$ and every non-empty closed $A\subseteq X\setminus\{x\}$ (or every closed $A\subseteq X\setminus\{x\}$, if you adopt the version with $f(A)\subseteq\{0\}$). Not just one closed set $A$. So, you could not say, oh I found an $f$ that works for $A=\varnothing$, and I don't have to worry about other closed $A$. (That is, we should consider all $x\in X$, and all (empty or non-empty, depending on which version of the definition you adopt) closed sets $A\subseteq X\setminus\{x\}$, and find a suitable $f$ for each such $x$ and $A$.)

  • $\begingroup$ Dear Mirko! How are you? Thanks a lot for your help! I appreciate it! $\endgroup$ – ZFR Oct 26 '19 at 18:54
  • $\begingroup$ @ZFR You are welcome, thank you! $\endgroup$ – Mirko Oct 26 '19 at 19:53

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