# Lebesgue's topological dimension

I was reading the definition of dimension from the book: "Topology", Munkres, 2nd ed., and surely I didn't understand, but I wonder how $\mathbb{R}^2$ can have dimension 2.

Take the open sets $U_n=\{(x,y)\mid −\infty <x <\infty ,n−1<y<n+1 \}$ for every integer $n$. It covers the plane but its order is 2, so the dimension should be less than 2.

Shouldn't be the difinition with balls or "squares"?

Thank you.

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There's no notion of balls, let alone squares, in a general topological space. If you're interested in a notion of dimension of metric spaces based on balls, look up "Hausdorff dimension". – MartianInvader Sep 5 '12 at 0:05

From Wikipedia, with emphasis added: The covering dimension of a topological space $X$ is defined to be the minimum value of $n$ such that every finite open cover $\mathcal{A}$ of $X$ admits a finite open cover $\mathcal{B}$ of $X$ which refines $\mathcal{A}$ in which no point is included in more than $n+1$ elements. The dimension of $\mathbb{R}^2$ is not 1 because there are coverings other than the one you defined, e.g., ones by balls or squares that you alluded to, that witness that the dimension is $\ge 2$.