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?
 A: 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 $[-1,1]^2$.
