
Tychonoff space
I'm having difficulty with proving directly that every metric space is a Tychonoff space. I can show that every metric space is Hausdorff.
Let $\displaystyle (X,d)$ be a metric space, let $\displaystyle A$ be a closed subset of $\displaystyle X$, let $\displaystyle y \in XA$. Hint: Define $\displaystyle f: X \rightarrow I$ by $\displaystyle f(x)= min\{d(x,A)/d(y,A),1\}$. So, $\displaystyle f(y)=1$ and $\displaystyle f(x)=0$ for every $\displaystyle x \in A$. I am stuck on proving that this function is continuous. Anyone can help me?

Since if $\displaystyle f$ and $\displaystyle g$ are continuous then so is $\displaystyle \min (f,g)$ you only have to show that $\displaystyle x\mapsto d(x,A)$ is continuous. It's a consequence of the triangular inequality.

Thank you very much for your help. I did prove $\displaystyle x \mapsto d(x,A)$ continuous before. I guess now I just need to show that $\displaystyle min \{f,g\}$ is continuous if $\displaystyle f, g$ are continous, and this should be straightforward.