# Thread: Proving (-infinity, 0) is open

1. ## 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?

2. ## Re: Proving (-infinity, 0) is open

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

Does that work?

3. ## Re: Proving (-infinity, 0) is open

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

4. ## 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 $(-\infty, 0)$ so x is an interior point of $(-\infty, 0)$. Since x could be any point in the set, the set is open.

5. ## 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?

6. ## Re: Proving (-infinity, 0) is open

If $|r|<|x|$, it is possible that $r<0$, which you don't want. You want $0. Now, $(x-r,x+r)$ is correct. This is because $x<0$, so $x-r<0$. Since $0, $x+r = -|x|+r<0$. So, you have both endpoints less than zero.

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### (0 infinity) open set example

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