 | | From: | tsmith | | Subject: | analysis proof help | | Date: | Sun, 23 Jan 2005 22:11:21 -0800 |
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 | Hello,
Let A and B be nonempty bounded subsets of R and let S be the set of all
sums a+b where a in A and b in B.
a) Prove that Sup S = Sup A + Sup B ("Sup" means supremum).
What I have so far:
s0 = Sup S, a0 = Sup A, b0 = Sup B.
s0 >= a + b.
a+b <= a0 + b0
But I am having trouble relating s0 to a0 + b0.
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| | From: | bryant_j_j at yahoo.com | | Subject: | Re: analysis proof help | | Date: | 23 Jan 2005 23:13:06 -0800 |
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| s0>= a + b for all (a,b) in A x B. hence s0>=a0+b0=sup A + sup B (why?)
sup A >= a for all a in A & sup B>= b for all b in B. hence
sup A + sup B >= a + b = s for all s in S. hence sup A + sup B>=s0
(why?).
hence sup A + sup B = Sup S.
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| | From: | bryant_j_j at yahoo.com | | Subject: | Re: analysis proof help | | Date: | 23 Jan 2005 23:12:02 -0800 |
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| s0>=a+b for all any A, b on B. hence also s0>=a0+b0=Sup A + Sup B
Sup A+Sup B >= a + b, any a in A, and B in B
>= s , any s in S
>=s0
hence Sup A + Sup B = Sup S. QED.
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| | From: | bryant_j_j at yahoo.com | | Subject: | Re: analysis proof help | | Date: | 23 Jan 2005 23:11:50 -0800 |
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| s0>=a+b for all any A, b on B. hence also s0>=a0+b0=Sup A + Sup B
Sup A+Sup B >= a + b, any a in A, and B in B
>= s , any s in S
>=s0
hence Sup A + Sup B = Sup S. QED.
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| | From: | William Elliot | | Subject: | Re: analysis proof help | | Date: | Mon, 24 Jan 2005 01:22:39 -0800 |
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| On Sun, 23 Jan 2005, tsmith wrote:
> Let A and B be nonempty bounded subsets of R and let S be the set of all
> sums a+b where a in A and b in B.
S = A + B
> a) Prove that Sup S = Sup A + Sup B ("Sup" means supremum).
>
if a + b in A + B, then a in A, b in B,
a <= sup A, b <= sup B, a + b <= sup A + sup B
Thus for all x in A + B, x <= sup A + sup B,
so sup A + sup B is upper bound for A + B
giving
1) sup A+B <= sup A + sup B
Now if for all a in A, b in B, a + b <= x
ie if x is upper bound for A + B,
for all b in B, b <= x - a; so x - a is a upper bound for B.
Thus by definition of sup
sup B <= x - a
furthermore for all a in A,
a <= x - sup B
Thus
sup A <= x - sup B
sup A + sup B <= x
for all upper bounds x of A + B, but as sup A+B is an upper bound of A+B
2) sup A + sup B <= sup A+B
and by those two inequalities, we are done.
> What I have so far:
>
> s0 = Sup S, a0 = Sup A, b0 = Sup B.
>
> s0 >= a + b.
>
> a+b <= a0 + b0
>
> But I am having trouble relating s0 to a0 + b0.
>
>
>
>
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| | From: | bryant_j_j at yahoo.com | | Subject: | Re: analysis proof help | | Date: | 23 Jan 2005 23:12:24 -0800 |
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| s0>= a + b for all (a,b) in A x B. hence s0>=a0+b0=sup A + sup B (why?)
sup A >= a for all a in A & sup B>= b for all b in B. hence
sup A + sup B >= a + b >= s for all s in S. hence sup A + sup B>=s0
(why?).
hence sup A + sup B = Sup S.
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| | From: | bryant_j_j at yahoo.com | | Subject: | Re: analysis proof help | | Date: | 23 Jan 2005 23:09:26 -0800 |
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| s0>= a + b for all (a,b) in A x B. hence s0>=a0+b0=sup A + sup B (why?)
sup A >= a for all a in A & sup B>= b for all b in B. hence
sup A + sup B >= a + b >= s for all s in S. hence sup A + sup B>=s0
(why?).
hence sup A + sup B = Sup S.
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| | From: | bryant_j_j at yahoo.com | | Subject: | Re: analysis proof help | | Date: | 23 Jan 2005 23:14:27 -0800 |
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| s0>= a + b for all (a,b) in A x B. hence s0>=a0+b0=sup A + sup B (why?)
sup A >= a for all a in A & sup B>= b for all b in B. hence
sup A + sup B >= a + b >= s for all s in S. hence sup A + sup B>=s0
(why?).
hence sup A + sup B = Sup S.
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