Let us say a topological space $X$ has property $\Phi([0,1])$, provided for each two distinct points $a,b\in X$, exists a continuous map $f:X\to [0,1]$ (with $[0,1]$ the usual topology) s.t. $f(a)=0$, $f(b)=1$. I wonder whether in the above definition, the requirement that the range is $[0,1]$ is essential. If we change it to $\mathbb{R}$ (with the requirement $f(a)=0$, $f(b)=1$ not changed), is the corresponding property $\Phi(\mathbb R)$ equivalent to $\Phi([0,1])$?

In other words, my question is summarized as:

$X$ is a topological space s.t. for each two distinct points $a,b\in X$, exists a continuous map $f:X\to \mathbb R$ ($\mathbb R$ has the usual topology) with $f(a)=0$, $f(b)=1$. For each $a\ne b\in X$, does there always exist a continuous map $g:X\to [0,1]$ with $g(a)=0, g(b)=1$?

  • $\begingroup$ In some sense the key fact here is that $\Bbb R$ itself has your property $\Phi([0, 1])$. If you change $0$ and $1$ to just "two distinct points" or "two particular distinct points", then if $X$ has property $\Phi(Y)$ and $Y$ has property $\Phi(Z)$, then $X$ has property $\Phi(Z)$. (This should be straightforward to prove) $\endgroup$ Apr 12 at 10:58

1 Answer 1


Consider the map $h:\mathbb{R} \to [0,1]$ that sends:

  • $2n < x < 2n+1$ to $x-2n$ for $n \in \mathbb{Z}$.
  • $2n+1 < x < 2n+2$ to $2n+2-x $ for $n \in \mathbb{Z}$.
  • $2n$ to $0$ and $2n+1$ to $1$

This map is clearly continuous and sends $0$ on $0$ and $1$ on $1$. For a "visual" description, $h$ sends $\mathbb{R}$ on $[0,1]$ by going back and forth, making a u-turn at every integer.

Then you just need to consider $g=h\circ f$.


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