Proving (-infinity, 0) is open

• Sep 29th 2013, 03:09 PM
glambeth
Proving (-infinity, 0) is open
Hi, I've done most of the proof(I think), but I'm lost anyone willing to finish it off for me? I've spent a while on it and i'm stuck.

Let x be an element of (-infinity, 0) this implies x < 0 chose an r such that abs(r) < abs(x) let y be an element of (x -r, x+r).
This implies x-r<y<x+r which implies 0 - x + x > x - r < y now what exactly do I do to show y < 0 which would be the end goal, right?
• Sep 29th 2013, 03:23 PM
Plato
Re: Proving (-infinity, 0) is open
Quote:

Originally Posted by glambeth
Hi, I've done most of the proof(I think), but I'm lost anyone willing to finish it off for me? I've spent a while on it and i'm stuck.
Let x be an element of (-infinity, 0) this implies x < 0 chose an r such that abs(r) < abs(x) let y be an element of (x -r, x+r).
This implies x-r<y<x+r which implies 0 - x + x > x - r < y now what exactly do I do to show y < 0 which would be the end goal, right?

If $\displaystyle x\in (-\infty,0)$ then $\displaystyle x\in (2x,0)\subset (-\infty,0)$. $\displaystyle (2x,0)$ is an open set containing $\displaystyle x$.

Does that work?
• Sep 29th 2013, 04:13 PM
glambeth
Re: Proving (-infinity, 0) is open
Hm that makes sense, but how would I be able to get to that stage?
• Sep 29th 2013, 04:24 PM
HallsofIvy
Re: Proving (-infinity, 0) is open
What you did was perfectly good- the "r-neighborhood" of x, (x-r, x+ r) is a subset of $\displaystyle (-\infty, 0)$ so x is an interior point of $\displaystyle (-\infty, 0)$. Since x could be any point in the set, the set is open.
• Sep 29th 2013, 04:43 PM
glambeth
Re: Proving (-infinity, 0) is open
Hm, I think my part is wrong though. Looking at it i say choose an r such that abs(r) < abs(x) but then i say x - r < y < x + r so i end up saying 0 < y < 2x but isn't this false? Given y should be less than 0?
• Sep 29th 2013, 08:08 PM
SlipEternal
Re: Proving (-infinity, 0) is open
If $\displaystyle |r|<|x|$, it is possible that $\displaystyle r<0$, which you don't want. You want $\displaystyle 0<r<|x|$. Now, $\displaystyle (x-r,x+r)$ is correct. This is because $\displaystyle x<0$, so $\displaystyle x-r<0$. Since $\displaystyle 0<r<|x|$, $\displaystyle x+r = -|x|+r<0$. So, you have both endpoints less than zero.