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?
Best Answer
Right.
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 $A$.)