# Understanding equivalent condition for covering dimension

Let dim $$X$$ denote the Lebesgue covering dimension for a topological space $$X$$. Now a result in common books concerning dimension theory states the following:

If $$X$$ is a normal topological space, then

1) dim $$X\leq n$$

if and only if

2) every continuous function $$f : X → {\mathbb{R}}^{n+1}$$ can be approximated arbitrarily closely by continuous functions which do not contain the origin in their range.

I have clearly understood the definition of covering dimension of a space. I am not able to picture the result. I am not able to connect the two notions 1) and 2). Can anyone help get a better understanding with a basic example?

• Intuitively this is almost trivial: say, if you continuously map a line to the plane and accidentally hit the origin then you can perturb the map arbitrarily slightly and the origin will be missed. While if you, say, map to the plane a disk and the origin is in the image then you might have quite some way to go until the origin gets out of it – მამუკა ჯიბლაძე Oct 3 '18 at 13:50

This works best via the theorem on partitions (Theorem 7.2.15 in Engelking's General Topology): $$\dim X\le n$$ iff for every sequence $$(A_1,B_1)$$, ..., $$(A_n,B_n)$$, $$(A_{n+1},B_{n+1})$$ of pairs of disjoint closed sets there is a sequence $$(U_1,V_1)$$, ..., $$(U_n,V_n)$$, $$(U_{n+1},V_{n+1})$$ of pairs of disjoint open sets, with $$A_i\subseteq U_i$$ and $$B_i\subseteq V_i$$ for all $$i$$, such that $$\bigcap_{i=1}^{n+1}L_i=\emptyset$$, where $$L_i=X\setminus(U_i\cup V_i)$$.
Your condition implies $$\dim X\le n$$ by considering continuous $$f_i:X\to[-1,1]$$ with $$f_i[A_i]=\{-1\}$$ and $$f_i[B_i]=\{1\}$$; use a function $$g$$ that is closer than $$1/3$$ to $$(f_1,\ldots,f_n,f_{n+1})$$ and that avoids the origin to create the $$U_i$$ and $$V_i$$.
Conversely, given $$f$$ let $$\varepsilon>0$$ and consider the closed sets $$A_i=\{x:f_i(x)\le-\varepsilon\}$$ and $$B_i=\{x:f_i\ge\varepsilon\}$$; find the $$U_i$$, $$V_i$$ and $$L_i$$ and use Urysohn functions $$g_i$$ that satisfy $$g[A_i]=\{-\varepsilon\}$$, $$g_i[L_i]=\{0\}$$ and $$g_i[B_i]=\{\varepsilon\}$$ to change $$f$$ outside the union of the $$A_i$$s and $$B_i$$s so as to avoid the origin.
For a concrete case look at maps from the unit interval to the plane; these can be approximated uniformly by piecewise linear maps and it is always possible to change the latter so as to avoid the origin. This can even be done with a surjective map from $$[-1,1]$$ onto the square $$[-1,1]^2$$.