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Prove: If p has a neighborhood contained in A, then p is in the interior of A.

  • Thread starter Hodgey8806
  • Start date
  • #1
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Homework Statement


If p has a neighborhood contained in A, then p is in the interior of A.


Homework Equations


Int A = [itex]\bigcup[/itex]{C[itex]\subseteq[/itex]X:C[itex]\subseteq[/itex]A and C is open in X}
By the books definition, a neighborhood is open.

The Attempt at a Solution


Let C'[itex]\subseteq[/itex]A be a neighborhood of p contained in A
Then p[itex]\in[/itex][itex]\bigcup[/itex]{C[itex]\subseteq[/itex]X:C[itex]\subseteq[/itex]A and C is open in X}
p[itex]\in[/itex]Int A

Seems too simple, but I'm going through the book's definition of "neighborhood of p"--and open subset of X.
 
Last edited:

Answers and Replies

  • #2
6,054
390
You are thinking along the right lines. But C' as you chose it leads you nowhere. You have not proved that p is in any of C.
 
  • #3
145
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Should it state something like:
Let p[itex]\in[/itex]A,
There exists an open set C'[itex]\subseteq[/itex]A s.t. p[itex]\in[/itex]C'
Thus it's in the union of all open sets.

Would that fix the issue? Thanks!
 
  • #4
6,054
390
That works.
 
  • #5
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Excellent! Thank you for the help!
 
  • #6
6,054
390
You just need to put things in a logical sequence without any gaps. First there is a neighborhood V. It contains some open set C'. Which must necessarily be one of all those open sets. So p is in the union. Q. e. d. If you are not explicit about every small detail, your professor might become unhappy :)
 
  • #7
145
3
Lol, my professor is a stickler--but she is an excellent teacher! So she's only a stickler because she has covered the material. So hopefully I catch these things before a test---though we have a quiz coming up!

If you have time, do you mind helping me with my other post? I'm not quite sure I can stop where I did--though it makes sense to me!
 

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