Can a Perfect Set Contain an Open Subset?

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Suppose we have a perfect set E\subset\mathbb{R}^k. Is there an open set I\subset E?
 
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I'm rusty at this. However, I understand that a closed interval is a perfect set. Take the closed unit cube in Rk, drop all boundary points leaving an open unit cube.
 
Sometime's it's true (like mathman's example), and sometimes it's false. For example the Cantor set is a perfect set but contains no open interval inside of it
 
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Yeah, I just found out that. Thank you :)
 

Thread '2-sphere intrinsic definition by gluing disks' boundaries'

A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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