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?

Re: Proving (-infinity, 0) is open

Re: Proving (-infinity, 0) is open

Hm that makes sense, but how would I be able to get to that stage?

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 so x is an interior point of . Since x could be any point in the set, the set is open.

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?

Re: Proving (-infinity, 0) is open

If , it is possible that , which you don't want. You want . Now, is correct. This is because , so . Since , . So, you have both endpoints less than zero.