1. ## subset question

If I have that A is an open interval the from the definition I have that $\displaystyle (a-r,a+r)\subset A\subset\mathbb{R}$ can I say that $\displaystyle a+r\in A$?

thanks for any help

ps sorry about the first post I sent it from my phone and clearly something went wrong sorry

2. ## Re: subset question

Originally Posted by hmmmm
If I have the following $\displaystyle (-a,a)\subset A\subset\mathbb{R}$ can I say that $\displaystyle a\in A$?
It could be that $\displaystyle (-2a,a)=A$.

4. ## Re: subset question

Originally Posted by hmmmm
No again.
Consider $\displaystyle (-2,2)=A$ now clearly $\displaystyle (1-1,1+1)\subset A$ right?
Does $\displaystyle 1+1\in A~?$

5. ## Re: subset question

yeah it isn't, however we can always choose an r so that it is?

6. ## Re: subset question

Originally Posted by hmmmm
yeah it isn't, however we can always choose an r so that it is?
Look this is a waste of time.
Post a problem. A complete well formed problem.
Stop making us guess as to what you are trying to ask.

7. ## Re: subset question

If A is an open subset then $\displaystyle supA\not\in A$

proof

Assume that $\displaystyle \alpha=sup A\in A$.

Then as A is open from the definition we have that for $\displaystyle \alpha\in A\ \mbox{and}\ r>0$

$\displaystyle (\alpha-r,\alpha+r)\subset A$.

Contradiction- as we assumed that $\displaystyle \alpha=Sup A$ but $\displaystyle \alpha+r>\alpha\ \mbox{and}\ \alpha+r\in A$.

I was asking if I can always choose an r such that $\displaystyle \alpha+r\in A$?

Thanks for any help

8. ## Re: subset question

Originally Posted by hmmmm
If A is an open subset then $\displaystyle supA\not\in A$proof
Assume that $\displaystyle \alpha=sup A\in A$.
Then as A is open from the definition we have that for $\displaystyle \alpha\in A\ \mbox{and}\ r>0$
$\displaystyle (\alpha-r,\alpha+r)\subset A$.
Contradiction- as we assumed that $\displaystyle \alpha=Sup A$ but $\displaystyle \alpha+r>\alpha\ \mbox{and}\ \alpha+r\in A$.
I was asking if I can always choose an r such that $\displaystyle \alpha+r\in A$?
Well thank you for finally posting an intelligible problem.
It is well know that an open set in $\displaystyle \mathbb{R}$ cannot contain a maximal element.
Recall that between any two real numbers there is a real number.

9. ## Re: subset question

Is my proof of this statement then correct, specifically where I derived the contradiction and said $\displaystyle \alpha+r\in A$?

10. ## Re: subset question

Originally Posted by hmmmm
Is my proof of this statement then correct, specifically where I derived the contradiction and said $\displaystyle \alpha+r\in A$?
No you cannot say that.
But you can say $\displaystyle \left( {\exists s} \right)\left[ {\alpha < s < r} \right]$ thus $\displaystyle s\in A~.$