A topological space $X$ is self-similar if there is a proper subspace $Y \subsetneq X$ such that $Y \simeq X$ (homeomorphic).
Is $\mathbb{S}^2$ self-similar? I guess, no. But, how to prove that?
Self-similar spaces (I)
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0Do you know about homotopy? – 2017-02-24
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5If $\mathbb{S}^2$were self-similar, then it could be embedded inside the plane (that's because $\mathbb{S}^2$ without a point is homeomorphic to $\Bbb{R}^2$). Can you prove that this is impossible (it is a consequence of Borsuk-Ulam)? – 2017-02-24
1 Answers
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Suppose $f:\mathbb{S}^2 \rightarrow Y$ is a homeomorphism where $Y \subsetneq \mathbb{S}^2$. Pick $p \in \mathbb{S}^2 \setminus Y$. Then $\mathbb{S}^2\setminus\{p\} \simeq \mathbb{R}^2$ via stereographic projection (from $p$), say $h$.
But then $h \circ f: \mathbb{S}^2 \rightarrow \mathbb{R}^2$ would be 1-1, while Borsuk Ulam says that $p \in \mathbb{S}^2$ exists with $h \circ f(p) = h \circ f(-p)$, contradiction.