Show If a point A is in the interior, then it has a neighborhood contained in A.

In summary, the conversation discusses the proof of showing that if a point A is in the interior of a topological space X, then it has a neighborhood contained in A. The conversation also includes a clarification of the notation used and a request for help with another question regarding the equivalence of a bounded set with a closed ball.
  • #1
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Homework Statement


Let A be a topological space and let A[itex]\subseteq[/itex]X be any subset.
Show: If a point A is in the interior, then it has a neighborhood contained in A.


Homework Equations


Neighborhoods are defined to be open in my book.
Int(A) = [itex]\bigcup[/itex]{C[itex]\subseteq[/itex]A and C is open in X}


The Attempt at a Solution


Let p[itex]\in[/itex]Int(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}
So, [itex]\exists[/itex] an open set C' s.t. p[itex]\in[/itex]C' and C[itex]\subseteq[/itex]A
Q.E.D.
 
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  • #3
Thank you! Could you check my work backwards now?
I can say:

Let C'[itex]\subseteq[/itex]A be open and non-empty
Let p[itex]\in[/itex]C
Thus p[itex]\in[/itex][itex]\bigcup[/itex]{C[itex]\subseteq[/itex]X:X[itex]\subseteq[/itex]A and C is open in X}
Thus p[itex]\in[/itex]Int A
 
  • #4
That's correct. But you got to clean up your presentation. What's C' for example??

Hodgey8806 said:
Thank you! Could you check my work backwards now?
I can say:

Let C'[itex]\subseteq[/itex]A be open and non-empty
Let p[itex]\in[/itex]C
Thus p[itex]\in[/itex][itex]\bigcup[/itex]{C[itex]\subseteq[/itex]X:X[itex]\subseteq[/itex]A and C is open in X}
Thus p[itex]\in[/itex]Int A
 
  • #5
C' was meant to be one set that has that element and
I meant to type let p[itex]\in[/itex]C'
and I meant to write
p∈⋃{C⊆X:C⊆A and C is open in X}

Do I still need to clean up?
 
  • #6
Ah ok. I guess it's ok now!
 
  • #7
Thank you very much! If you have time, would you mind helping me with my other question? It's about the equivalence of a bounded set with a closed ball. I just want to check my proof one direction. Thanks again!
 

1. What does it mean for a point to be in the interior?

For a point to be in the interior, it means that the point is located within the boundaries of a given set or region.

2. What does it mean for a point to have a neighborhood?

A neighborhood of a point is a set of points that are in close proximity to the given point. It can be thought of as a small region or area surrounding the point.

3. How can we determine if a point is in the interior of a set?

To determine if a point is in the interior of a set, we can check if the point is contained within the boundaries of the set and also if it has a neighborhood contained within the set.

4. What does it mean for a neighborhood to be contained in a point?

A neighborhood being contained in a point means that all points within the neighborhood are also within the boundaries of the given point. Essentially, the neighborhood is a subset of the point.

5. How does this statement relate to topology?

This statement is a fundamental concept in topology, as it helps define the interior of a set and the properties of points within the set. It is used to determine open sets, connectedness, and other topological properties of a given set.

Suggested for: Show If a point A is in the interior, then it has a neighborhood contained in A.

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