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

[Shackleford]

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

 

[gauss^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 x \in (a,b) (assuming a < b) and take r to be the minimum of b-x and x-a so that x - r >= x - (x-a) = a and x + r <= x + (b-x) = b. Then (x-r, x+r) \subset (a,b) holds so that x is an interior point of (a,b).

 

[Shackleford]

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 x \in (a,b) (assuming a < b), take r to be the minimum of b-x and x-a so that x - r >= x - (x-a) = a and x + r <= x + (b-x) = b. Then (x-r, x+r) \subset (a,b) holds so that x is an interior point of (a,b).

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.