Topology, Int(A) is an open set

  • Thread starter rourky
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  • #1
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Homework Statement



Question: Prove Int(A) is an open set, given Int(A) is the set of all interior pts of A where x is an interior pt of A if it is the centre of an open ball in A.

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The Attempt at a Solution



Attempted Soln: Suppose x is an element of Int(A).
Then there exists r > 0 such that B(x, r) is a subset of A.
Have tried to extend this to say there exists r > 0 such that B(x, r) is a subset of Int(A), but with no success.

Have also tried to prove C(Int(A)) contains all its limit pts and thus is closed. Then Int(A) would be open.

Just looking for a hint to get me on the right road. Thanks
 

Answers and Replies

  • #2
Dick
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Take a point y in your B(x,r). Can you also show y is in Int(A)? Hint: consider |x-y|, can you say anything about that distance?
 
  • #3
HallsofIvy
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By the way, there are a number of different ways of approaching topology. You appear (since you say "there exists r > 0 such that B(x, r) is a subset of Int(A)") to be assuming a topology defined by a metric. I notice Dick refers to |x-y| rather than d(x,y), so he may be assuming you are in R, the real numbers with the usual topology. In more general topology an "open set" is simply a member of the "topology" and the "interior" of a set, A, is defined as the union of all open subsets of A. It would help if you would say explicitely what kind of topology you have, what definitions you are using. In any case, since you are definitely in a metric space, you can use the "triangle" inequality: that [itex]d(x,y)\le d(x,z)+ d(z,y)[/itex] (or [itex]|x-y|\le |x-z|+ |z- y|[/itex]. That should be useful.
 
  • #4
Dick
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Well, I wasn't so much assuming anything as simply being sloppy. As Halls points out, you don't even really need the metric. Since B(x) is open, if y is an element of B(x) then___? Fill in the blanks.
 
  • #5
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Thanks for your replies guys, problem solved! Yes, I was assuming a topology by a metric, sorry for not stating so.
 

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