# Definition of Tychonoff space or $T_{3\frac{1}{2}}$ space

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?

• 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 ($\forall x \in A: f(x)=0$).. – 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 $$A$$.)