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Does an open subset of $R^n$ exist that has dimension less than $n$ in the standard topology? All the less than $n$ dimensional subsets I can think of are not open.

MeMyselfI
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    Your hint is that any ball in $\mathbb R^n$ has non-zero width in all the dimensions, hence can't be contained in any specific smaller dimension. – Sarvesh Ravichandran Iyer Apr 09 '17 at 07:07
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    There're a lot of different dimension notions, but most of them are local in nature, and as topology on $\mathbb R^n$ is generated by balls (think of: every point in open set contains some neighbourhood), every open set will have same dimension as $\mathbb R^n$ itself, as any ball is homeomorphic to it. – xsnl Apr 09 '17 at 07:11

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In $\mathbb{R}^n$, open balls are homeomorphic to the whole space (use $f(x) = \frac{1}{1+||x||}\cdot x$, for $\mathbb{R}^n$ and the open unit ball, and use translations and scaling to make all open balls homeomorphic among themselves.) This implies $\dim(B) = \dim(\mathbb{R}^n) = n$ (the latter is a deep theorem)

and dimension functions are monotone in metric spaces, so if a set $A$ contains some open ball $B$, then $\dim(B) \le \dim(A)$. So all open sets have dimension $n$.

Henno Brandsma
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