# Is the following two statement on a topological space equivalent?

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 $$g(a)=0, g(b)=1$$?

• 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) Apr 12 at 10:58

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 $$g=h\circ f$$.