Thread: Proving open sets

A set S of real numbers is defined to be an open set if it has the following property: for each xEs, there exists a positive number r such that (x-r, x+r) is a subset of S. The set S is closed if the set R(real numbers)\S is open.

Prove the interval (a,b) is an open set and prove the interval (a, infinity) is an open set.

Prove the interval (a,b) is an open set

Let b . Take . Then, .

Now, try to prove that the other set is open.

Prove the interval (a,infinity) is an open set

Let xE(a,infinity). Take r=max(|x-a|,|x-infinity|) then (x-r, x+r) is a subset of (a,infinity) for all x which is an element if the set (a,infinity).

Would this be right?

you cant do things such as |x-infinity|, infinity is not a number.

Using the proof that felper gladly posted, understand the idea, and use that ide in the case the set is (a, infinity)



Remember: there is no such thing as 6-infinity. That is not defined.

Originally Posted by summerset353

Prove the interval (a,infinity) is an open set

Let xE(a,infinity). Take r=max(|x-a|,|x-infinity|) then (x-r, x+r) is a subset of (a,infinity) for all x which is an element if the set (a,infinity).

Would this be right?

As mabruka said, |x- infinity| is non-sense. Also felper suggested you use "min", not "max". Why did you change?

Whatever meaning we might want to assign to "|x- infinity|", perhaps in a limit sense, it is clearly going to be larger than |x- a| so min(|x-a|, |x- infinity|), having no meaning, can be replaced by just |x-a|.

Originally Posted by summerset353

A set S of real numbers is defined to be an open set if it has the following property: for each xEs, there exists a positive number r such that (x-r, x+r) is a subset of S. The set S is closed if the set R(real numbers)\S is open.

Prove the interval (a,b) is an open set and prove the interval (a, infinity) is an open set.

According to the above definition to prove (a,b) is open:

Let xε(a,b) => a<x<b => x-a>0 and b-x>0.

Choose r=min{ x-a, b-x} => r<x-a and r<b-x.

Now to prove (x-r,r+x) (a,b).

So, let yε(x-r,x+r) => x-r<y<x+r => a-x+x< x-r<y<x+r<b-x+x => a<y<b => yε(a,b)

Now to prove is open :

Again,let xε => x>a => x-a>0

Choose r<x-a

And to prove (x-r,x+r) ,

let yε(x-r,x+r) => x-r<y<x+r => a-x +x<x-r<y< x+r< 2x-a => y>a => yε