Show that every interior point of a set must also be an accumulation point of that set, but not conversel
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Any point x that belongs to E is said to be an interior point of E provided that some interval (x−c, x+c)⊂E.
Any point x (not necessarily in E) is said to be an accumulation point of E provided that for every c>0 the intersection (x−c, x+c)∩E contains infinitely many points.
How do I show that every interior point of a set must also be an accumulation point of that set from these 2 definitions?
real-analysis
general-topology
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edited Oct 12 '16 at 20:08
BCLC
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asked Oct 12 '16 at 17:17
Siddhartha
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4 Answers
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Let x be an interior point of E, (x−c,x+c)⊂E. Let d>0, (x−inf(c,d),x+inf(c,d))∩E⊂(x−c,x+c)⊂E. Since (x−inf(c,d),x+inf(c,d))⊂(x−d,x+d) contains an infinite number of points, the result follows.
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answered Oct 12 '16 at 17:21
Tsemo Aristide
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There is ε>0 such that (x−ε,x+ε)⊂E. Let εn=εn. Take xn∈(x−εn,x+εn)∖(x−εn+1,x+εn+1). The set {xi}∞i=1∩(x−c,x+c) is infinite for any c>0.
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edited Oct 12 '16 at 17:25
answered Oct 12 '16 at 17:21
Matias Heikkilä
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It would be clearer to write {x1,x2,…} where you have written {xn} (which looks like a singleton set at first glance, forcing the reader to look at the context to figure out what you actually meant). – Rob Arthan Oct 12 '16 at 17:24
Thank you for your comment. I changed the notation. – Matias Heikkilä Oct 12 '16 at 17:25
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Let x be an interior point of E. Since x is an interior point of E there exists c>0 such that (x−c, x+c)⊂E. Now, you have to show that it is an accumulation point. That is, you have to show that for any d>0 the set (x−d,x+d)∩E contains infinitely many points. Of course, it is enough to consider d≤c. But then (x−d,x+d)∩E=(x−d,x+d) (since (x−c, x+c)⊂E) which clearly contains infinitely many poitns. (If d>c use that (x−c,x+c)⊂(x−d,x+d)∩E.)
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