A topological space $X$ is said to be Tychonoff if for each compact subset $S\subset X$ and for all $x\in X\setminus S$ there exists a continuous function $f:X\longrightarrow[0,1]$ such that $f(x)=1$, and $f|_S=0$.

I guess it is a widely known fact. However... Let $X$ be a Tychonoff space, $K\subset X$ - a compact subspace. Is it true that there exists a continuous function $f:X\to [0,1]$ such that $f|_K=0$ but $f(x)\ne 0$ for all $x\notin K$?

  • $\begingroup$ Do you mean "$f(x) \not = 0$ for all $x\not\in K$" at the end? $\endgroup$ – anomaly Jun 18 '14 at 16:08
  • $\begingroup$ Yes, I do. Thanks. $\endgroup$ – Kolyan Jun 18 '14 at 16:09

No. Consider the ordinal space $\omega_1 + 1 = [ 0 , \omega_1 ]$. This is a compact Hausdorff space (hence Tychonoff, and even normal). Note that the singleton $\{ \omega_1 \}$ is a closed (compact) subset, but there is no continuous function $f : [0,\omega_1] \to [0,1]$ such that $f(\omega_1) = 0$ and $f(\xi) \neq 0$ for all $\xi < \omega_1$. (If $f$ were such a function, then for each $n > 0$ there is a $\alpha_n < \omega_1$ such that $f(\xi) < \frac{1}{n}$ for all $\alpha_n < \xi \leq \omega_1$, but if you consider $\alpha = \sup_{n} \alpha_n$, it can be shown that $\alpha < \omega_1$, but then $f(\xi) = 0$ for all $\alpha \leq \xi \leq \omega_1$.)

(I explain a bit more in this previous answer.)


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