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  1. $A$ and $B$ bounded sets of reals implies $D=\{xy|x \in A, y \in B\}$ bounded.
  2. $A$ a bounded set of reals, $0 \notin A$ implies $E=\{1/x|x \in A\}$ bounded.
  3. $A$ bounded set of reals implies $F=\{x^{5} | x \in A\}$ bounded.

For all three, my gut tells me they are all true. Aren't these just operations to the elements of a bounded set? I can't see any way these situations produce an unbounded set. Am I missing something?

Thanks!

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2 Answers 2

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The 2nd one is false.

Anyway, when confronted with problems of this kind you should always prove the affirmations you think are true and find counterexamples for the ones you think are false. In this case you have to find bounds for the sets $D$ and $F$ (they will depend on $A$ and $B$, naturally).

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oh ok I think I see, so for example for set D. Would the upper bound be lub(A)*lub(B) and the lower bound be glb(B)*glb(A)? –  Peej Gerard Feb 3 '13 at 19:29
    
sorry those would be the glb and lub of D actually right? But having a glb and lub would imply that there exists upper and lower bounds for the set D though? –  Peej Gerard Feb 3 '13 at 19:34
    
Yes, if you have an upper and a lower bound for your set, then it is bounded (that's the definition). But you have to be careful. For example if $A = [-2,-1]$ and $B = [-5,3]$ then $D = [-6,10]$ and the lowest upper bound for $D$ would be 10, which is different from the product of the lowest upper bound of $A$ times the one of $B$. Try to think a little bit more about it. You're on the right track. –  Daniel Robert-Nicoud Feb 3 '13 at 20:32

The second one is false. Put $A = (-1,1) - \{0\}$.

To elaborate, for any $n$, we have $1/n\in A$, so $n \in E$.

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What have you tried on 1 and 3? –  ncmathsadist Feb 3 '13 at 19:23
    
hmm, ok can you elaborate on why this would be unbounded? –  Peej Gerard Feb 3 '13 at 19:23
    
is it because there are infinite real values between say (-1,1)? –  Peej Gerard Feb 3 '13 at 19:25
    
Correct. You can come as close to zero as you wish, so the reciprocal can be made arbitrarily large. –  ncmathsadist Feb 3 '13 at 20:17

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