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3 votes
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Is every weakly $1$-dimensional space embeddable in the plane?

A $1$-dimensional (separable metric) space $X$ is weakly $1$-dimensional if $$\Lambda(X)=\{x\in X:X\text{ is 1-dimensional at }x\}$$ is zero-dimensional (i.e. the space $\Lambda(X)$ has a basis of ...
D.S. Lipham's user avatar
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5 votes
1 answer
198 views

Iterating the dimensional kernel of a metric space

Fix $n\in \mathbb N$. Let $X$ be a separable metric space of (inductive) dimension $n$. Let \begin{align} \Lambda(X)&=\{x\in X:X\text{ is $n$-dimensional at }x\}\\ \\ \Lambda^2(X)&=\Lambda(\...
D.S. Lipham's user avatar
  • 3,317
6 votes
0 answers
111 views

A generalized Hausdorff dimension in form of a Lower semi continuous function

Let $(X,d)$ be a compact metric space. Assume that $f:X\to \mathbb{R}$ is a positive continuous function. We say that the $f$-dimension of $(X,d)$ is equal to $0$ if for every $\epsilon>0$ ...
Ali Taghavi's user avatar