Is There a Shape with More than 2^{\aleph_0} Sides in the Euclidean Plane?

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The discussion explores the possibility of shapes in the Euclidean plane having more than 2^{\aleph_0} sides, suggesting that such shapes might be fractals. It emphasizes the need to define "side," proposing that if a side is an ordered pair of vertices, then more than 2^{\aleph_0} unique vertices would be necessary. The conversation highlights that achieving this in the x-y plane appears impossible, potentially requiring higher-dimensional spaces. The participants acknowledge that while the x-y plane may be limiting, there are R^R choices for possible shapes. Ultimately, the topic raises intriguing questions about geometry and dimensionality in mathematical spaces.
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Do you think it is possible to have a shape in the euclidean plane that has more
than 2^{\aleph_0} sides. maybe some crazy fractal.
and if not in the plane would we maybe have to go to infinite dimensional space.
 
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Define "side".

Anyway, we have that \{(x,y)~\vert~x,y\in \mathbb{R}\} has cardinality 2^{\aleph_0}. So your figure would have to repeat its "sides" infinitely often.
 
micromass said:
Define "side".

Well, if a "side" is an ordered pair of vertices then you'd need more than 2^{\aleph_0} possible values for each vertex in order to have more than 2^{\aleph0} possible ordered pairs of vertices.

That sounds like you need a space whose dimensionality is larger than the cardinality of R.
 
thanks for the responses. defining a side with an ordered pair seems like a good idea
you it seems like you couldn't do it in the x-y plane. it does seem like I
have R^R choices though for possible shapes.
 
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