If I is open interval, prove I is an open set

In summary, the conversation is about proving a mathematical problem using neighborhoods. The speaker suggests using neighborhoods to show the proof, but the other person argues that it would be simpler to take a value and set a radius to prove the same result. The professor also mentions that using the fact that one set is open and its complement is closed may not be enough to prove the problem.
  • #1
Shackleford
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2
Is this a good-enough proof? I could have used neighborhoods to show this, but it seems like this way is a bit easier.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110911_113143.jpg [Broken]
 
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  • #2
You have really only changed the problem from showing that one set is open to showing that two are closed; you'll need to use neighborhoods either way. Much simpler to just take [itex]x \in (a,b)[/itex] (assuming [itex]a < b[/itex]) and take [itex]r[/itex] to be the minimum of [itex]b-x[/itex] and [itex]x-a[/itex] so that [itex]x - r >= x - (x-a) = a[/itex] and [itex]x + r <= x + (b-x) = b[/itex]. Then [itex](x-r, x+r) \subset (a,b)[/itex] holds so that [itex]x[/itex] is an interior point of [itex](a,b)[/itex].
 
  • #3
gauss^2 said:
You have really only changed the problem from showing that one set is open to showing that two are closed; you'll need to use neighborhoods either way. Much simpler to just take [itex]x \in (a,b)[/itex] (assuming [itex]a < b[/itex]), take [itex]r[/itex] to be the minimum of [itex]b-x[/itex] and [itex]x-a[/itex] so that [itex]x - r >= x - (x-a) = a[/itex] and [itex]x + r <= x + (b-x) = b[/itex]. Then [itex](x-r, x+r) \subset (a,b)[/itex] holds so that [itex]x[/itex] is an interior point of [itex](a,b)[/itex].

Well, I also used the fact that if one set is open, its complement is closed. I suppose that's not enough, though.

The professor suggested the method you just did.
 

What does it mean for a set to be open?

For a set to be open, it means that every point in the set has a small enough neighborhood around it that is still contained within the set. In other words, there are no points on the boundary of the set that are considered part of the set.

What is an open interval?

An open interval is a set of real numbers that includes all numbers between two given values, but does not include the endpoints. For example, the interval (0,1) would include all numbers between 0 and 1, but not 0 or 1 themselves.

How is the openness of a set related to its intervals?

If a set is open, it means that for every point in the set, there is an open interval containing that point that is still within the set. This is because the definition of an open set is based on the concept of open intervals.

Why is it important to prove that an open interval is an open set?

Proving that an open interval is an open set is important because it helps to establish the properties and characteristics of open sets. It also helps to build a foundation for understanding more complex concepts and theorems in mathematics and other scientific fields.

How can you prove that an open interval is an open set?

To prove that an open interval is an open set, you can use the definition of an open set and show that for every point in the interval, there exists an open interval containing that point that is still contained within the original interval. This can be done through logical reasoning and using the properties of real numbers.

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