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analysis PLEASE HELP!

i am trying to prove that

sup(S1 U S2 )=max{sup(S1 ),sup(S2 ).
where S1 and S2 are subsets of R.

so....

there is some number b such that for all z in (S1 U S2 ) b»z.

Now z is an element of (S1 U S2 ) implies that z is an element of S1 or Z is an element of S2 .

These are my ideas.

Anyone like to continue or suggest something else?...
Reply 1
You might like to prove

(1) max{sup(S1),sup(S2)} is an upper bound for S1uS2

(2) if z < max{sup(S1),sup(S2)} then z is not an upper bound for S1uS2
Reply 2
Jonny W
You might like to prove

(1) max{sup(S1),sup(S2)} is an upper bound for S1uS2

(2) if z < max{sup(S1),sup(S2)} then z is not an upper bound for S1uS2


(1) is what i seem to be having trouble with...its what i was trying to prove before..

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Jonny W
You might like to prove

(1) max{sup(S1),sup(S2)} is an upper bound for S1uS2

(2) if z < max{sup(S1),sup(S2)} then z is not an upper bound for S1uS2


(1) is what i seem to be having trouble with...its what i was trying to prove before..
any hints on this bit? xxx
Reply 3
Suppose that x \in S1uS2. Then (i) x \in S1, or (ii) x \in S2 (as you said).

In case (i), x <= sup(S1), so x <= max{sup(S1), sup(S2)}.

In case (ii), x <= sup(S2), so x <= max{sup(S1), sup(S2)}.


So x <= max{sup(S1), sup(S2)}.

ie, max{sup(S1), sup(S2)} is an upper bound for S1uS2.
Reply 4
Jonny W
Suppose that x \in S1uS2. Then (i) x \in S1, or (ii) x \in S2 (as you said).

In case (i), x <= sup(S1), so x <= max{sup(S1), sup(S2)}.

In case (ii), x <= sup(S2), so x <= max{sup(S1), sup(S2)}.


So x <= max{sup(S1), sup(S2)}.

ie, max{sup(S1), sup(S2)} is an upper bound for S1uS2.


cool thanks for that. i will try and put it all together now!
You haven't yet stated that max{sup(S1), sup(S2)} is in fact the *least* upper bound of S1US2...

To finish off the proof:

let y < max{sup(S1), sup(S2)}

by aritrary choice:
if max{sup(S1), sup(S2)} = sup(Si)
y < sup(Si)

suppose y is an upper bound of S1US2

there does not exist an element of S1US2 > y
however S1US2 contains all elements of Si
therefore there does not exist an element of Si > y

y is therefore an upper bound of Si, but y < sup Si by our choice. Contradiction.

You can choose i = 1 or 2 arbitrarily


QED

Edit: Actually just realised how similar our strategies are... you could probably work them in together
Reply 6
Olek
You haven't yet stated that max{sup(S1), sup(S2)} is in fact the *least* upper bound of S1US2...

To finish off the proof:

let y < max{sup(S1), sup(S2)}

by aritrary choice:
if max{sup(S1), sup(S2)} = sup(Si)
y < sup(Si)

suppose y is an upper bound of S1US2

there does not exist an element of S1US2 > y
however S1US2 contains all elements of Si


could you not just stop here and say that by our choice, y<sup(Si ) which in itself is a contradiction (we have something bigger than y which is an element of S1 U S2 ? xxx :rolleyes:

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