Question about nowhere dense sets

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The discussion centers on the properties of nowhere dense sets within the context of topological vector spaces (TVS). It establishes that if a subset A of a TVS X has a nonempty interior, then for any non-zero scalar k, the interior of kA is also nonempty. This conclusion relies on the fact that scalar multiplication in a TVS acts as a homeomorphism, preserving the topological structure. The participants agree that this property holds true, affirming the scale invariance of the topology in TVS.

PREREQUISITES
  • Understanding of topological vector spaces (TVS)
  • Familiarity with the concept of interior points in topology
  • Knowledge of homeomorphisms and their properties
  • Basic principles of metric spaces
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  • Research the properties of homeomorphisms in topological vector spaces
  • Study the implications of scale invariance in topology
  • Explore examples of nowhere dense sets in various metric spaces
  • Learn about the relationship between scalar multiplication and topology in vector spaces
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Suppose you know k_0A, for some set A \subset X (where X is a metric space) and some constant k_0, has nonempty interior. Do you then know that A has nonempty interior, and/or that k A has nonempty interior for any constant k?
 
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I don't know what you mean by "k_0A". Multiplication isn't defined in a general metric space.
 
HallsofIvy said:
I don't know what you mean by "k_0A". Multiplication isn't defined in a general metric space.

Point taken. Suppose we are in a vector space on which a metric has been defined.
 
I have strong feeling the following result holds in any topological vector space X:

Let A be subset of X, and k a non-zero scalar. Then k\cdot\text{int}A=\text{int}kA.

(Use that multiplication with k is a homeomorphism. Basically, this means the topology in a TVS is scale invariant. Haven't worked out the details.)

Assuming this is true, the answer to both your questions is then 'yes'.
 

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