Another neighborhood proof

Grant Bonner

New member
Joined
Aug 27, 2009
Messages
13
Ok, here's the deal. My friend and I are having an arguement on how to best prove the following.

Suppose x is a real number and e>0. Prove that (x-e,x+e) is a neighborhood of each of its members: in other words, if y is an element of (x-e,x+e) then there is d >0 such that (y-d,y+d) is a subset of (x-e,x+e) I will use P as a neighborhood of x, and Q as a neighborhood of y.

I think e<(d/2) as to make sure P is not contained in Q. 2d<=??????<2e<d. I also though about assuming z was an element of Q-P. I've got much scratch work, but nothing solid to start on.

She thinks it should be proof by case. i.e. d<e, d=e, d>e.

We both agree for the statement to be true d <e, or d=e and x=y.

This may not make sense. My head hurts.

I'm leaning towards proof by contradiction because it seems to be something we use heavily in class and we haven't done a proof by case yet.
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
11,329
Suppose that \(\displaystyle y \in \left( {x - \varepsilon ,x + \varepsilon } \right)\).
Let \(\displaystyle \delta = \min \left\{ {\left| {x - \varepsilon - y} \right|,\left| {x + \varepsilon - y} \right|}\right\}\).
Then \(\displaystyle \left( {y - \delta ,y + \delta } \right) \subseteq \left( {x - \varepsilon ,x + \varepsilon } \right)\).
 

daon

Senior Member
Joined
Jan 27, 2006
Messages
1,284
You may also use \(\displaystyle \delta = \epsilon-|x-y|\).

That quanity is always positive if y is different from x (why?). It is also equivilant to pka's suggestion.
 
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