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View Full Version : What does C^r-close mean?


quasar987
Jan19-11, 10:16 AM
For instance, if one says that a surface S embedded in a 3-manifold is C^{\infty}-close to another surface S', what does that mean?

Tinyboss
Jan22-11, 11:19 AM
(Just guessing, as I couldn't google up anything, either.)

If S and S' were embeddings of the same surface, and if for each p in the domain, you had a chart on your 3-manifold containing S(p) and S'(p), then I guess you could require the mixed partial derivatives of all orders to be close in magnitude.

Would that make any sense in context?

quasar987
Jan22-11, 11:33 AM
Yeah it would... and I found a nice topological way to express what you said:

A statement such as "If S=f(S_0) and S'=g(S_0) are C^r close to each other, then P." must be interpreted to mean "There exists a neigborhood U of f in C^r(S_0,M³) with the Whitney strong topology such that for all embeddings g in U, property P concerning f(S_0) and g(S_0) holds." (And embeddings are dense in C^r(S_0,M³) with the Whitney strong topology).