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.