Convex Subsets of Topological Vector Spacesby Edwin Tags: convex, spaces, subsets, topological, vector 

#1
Feb2409, 11:40 PM

P: 167

I had a quick question:
Is the following proof of the theorem below correct? Theorem: If C is a convex subset of a Topological vector space X, and the origin 0 in X is contained in C, then the set tC is a subset of C for each 0<=t<=1. Proof: Since C is convex, then t*x + (1t)*y is contained in C, for every x,y in C, and for 0<=t<=1. Since y = 0 is contained in C, then in particular, t*x = t*x + (1t)*0 = t*x + (1t)*y is contained in C, for every x in C, and for 0<=t<=1. Hence tC is a subset of C for 0<=t<=1. This completes the proof. Is the above proof correct? Or, did I make a mistake in the proof? 



#2
Feb2509, 01:21 AM

P: 322

Yes, the proof is correct. As a side remark, note that you did not need the vector space to be equipped with a topology. Thus the theorem is true (with the same proof) for any vector space over the real or complex numbers.




#3
Feb2509, 02:59 PM

P: 167

Thank you for the information! I had not noticed that.
So then the essential information in the proof are the vector space axioms (where the scalar field is either the real or complex numbers), and the definition of a convex set. The case of a Topological vector space is then just a special case of the more general theorem applied to an arbitrary vector space. Interesting. 


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