General topology: Prove a Set is Open

[lep11]

It's just the triangle inequality again:

$d(0, a) \le d(0, z) + d(z, a)$

$d(0, a) \le d(0, z) + d(z, a)$ ⇔ $d(0, z) ≥ d(0, a)-d(z, a)$

 

[PeroK]

$d(0, a) \le d(0, z) + d(z, a)$ ⇔ $d(0, z) ≥ d(0, a)-d(z, a)$

That's true,but what about using the inequalities you know for $d(0, a)$ and $d(z, a)$?

 

[lep11]

$d(0, z) ≥ d(0, a)-d(z,a) >d(0,a)>1$

 

[PeroK]

$d(0, z) ≥ d(0, a)-d(z,a) > d(0,a)>1$

That middle equality cannot be correct. $d(0, a)-d(z,a) \le d(0,a)$ surely?

 

[lep11]

That middle equality cannot be correct. $d(0, a)-d(z,a) \le d(0,a)$ surely?

What's wrong?

 

[PeroK]

What's wrong?

Come on! If you take a positive number away what you have gets smaller.

 

[lep11]

Come on! If you take a positive number away what you have gets smaller.

Okay, true.

I might just give up. I am not smart enough to study crap like this.

 

[lep11]

Okay, true.

I might just give up. I am not smart enough to study crap like this.

I took a break and tried again.

$d(0,z)≥d(0,x)-d(x,z)≥r+1-d(x,z)>r+1-r=1$

 

[lep11]

Thank You PeroK!