Tychonoff space

**jackie** · May 14th 2011, 10:24 AM

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

**girdav** · May 14th 2011, 10:30 AM

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

**jackie** · May 14th 2011, 10:44 AM

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

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