For example, if we have sets $\mathbb{S}^2$ and $G=[0,1] \times [1,2]$, those are obviously contained in two different dimensions, are they homeomorphic? I can't find any reason why they shouldn't be. Thank you.
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"Dimension" is not something you can define for a general topological space. The topology can't tell the difference in dimension. On the other hand, nonempty open sets $U \subseteq \mathbb{R}^n$, $V \subseteq \mathbb{R}^m$ with the subspace topology are not homeomorphic for $m \neq n$, although I think this is not so easy to prove; see here – Jair Taylor Mar 24 '19 at 23:37
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1You can define a dimension on any topological space. – H1ghfiv3 Mar 25 '19 at 06:30
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Those particular sets aren't homeomorphic, but there's no reason sets "contained in two different dimensions" can't be. The plane $\mathbb{R}^2$ is homeomorphic to the plane $x+y+z=0$ in $\mathbb{R^3}$, or to the hemisphere $\{(x,y,z)\mid x^2+y^2+z^2 =1\text{ and }z>0\}$
An object $X$ being embedded in some larger space $Y$ is not a topological property of $Y$, so homeomorphism won't respect that.
jmerry
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$\mathbb{S}^2$ and $[0,1]\times [1,2]$ can not be homeomorphic, because one of them is retractable and the other one is not.
Def: A topological space is called retractable, if you can deform it continuously to a point (by staying inside the space and without making holes). More rigorously, it is called retractable, if there is no homotopy between the identity function and a constant function.
Strichcoder
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