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I'm trying to define a real circle/disk without reference to angle, or distance. Is it characterized by the properties in the title?

If so, is it possible to then prove the Pythagorean theorem for this shape?

To be clear, by symmetric I mean that for every vector $$v \in D \iff -v \in D$$.

etha7
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    What does "symmetric in every basis" mean? – Theo Bendit Apr 04 '19 at 04:14
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    It's worth pointing out that the unit disc is an artefact of the dot product. If you change to a different inner product, you'll get a different unit disc (specifically, some kind of ellipse), but you'll still have Pythagoras's Theorem hold (and indeed, there will be a distance/angle preserving isomorphism between this space and $\Bbb{R}^2$ under the dot product). – Theo Bendit Apr 04 '19 at 04:17
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    Surely the unit square also has this property. Also, I'm not sure how you're going to get uniqueness without distance. How do you distinguish the unit circle from the circle of radius 2? – eyeballfrog Apr 04 '19 at 04:22
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    I do not see what your definition of symmetric has to do with a basis – Connor Malin Apr 04 '19 at 04:22
  • @eyeballfrog I guess I'd settle for an equivalence class of circles. Or I'd say something like a circle of radius r is the minimal circle containing $$(0,r)$$. – etha7 Apr 04 '19 at 04:28

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Given a convex, centrally symmetric set $D$, you want to know if $D$ is necessarily the unit disk. Well, if $D$ satisfies some other topological properties (compact with nonempty interior), then $D$ is indeed the unit disk... of some norm. There are lots of different norms, and there are lots of different possible shapes for $D$. See the answers to this question: What polytopes can be induced by a norm?

Chris Culter
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