Problem with supremum and infimum examples

  • #1

Main Question or Discussion Point

So I've got a calculus test in a week, and I'm studying for it but I can't understand some examples our professor has given us. So, he says:

1) A = { x[itex]\in[/itex] ℝ: (x-a)(x-b)(x-c) < 0 } , a<b<c. Find the supA and infA.

In the solution of his example he says. It is easy to see that A = (-∞,a)[itex]\cup[/itex](b,c) so infA = -∞ and supA = c. How did he find that A = (-∞,a)[itex]\cup[/itex](b,c) ?

2) B = {1 +(-1)n:n[itex]\in[/itex] N}. Find the supB and infB.

In the solution he says. Obviously B={0,1} which we can compute if we make n=2k (even) and n=2k+1 (odd)
so infB=0 and supB=1. So if n=2k wouldn't B be 2? Is this a mistake my professor made?
 

Answers and Replies

  • #2
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So I've got a calculus test in a week, and I'm studying for it but I can't understand some examples our professor has given us. So, he says:

1) A = { x[itex]\in[/itex] ℝ: (x-a)(x-b)(x-c) < 0 } , a<b<c. Find the supA and infA.

In the solution of his example he says. It is easy to see that A = (-∞,a)[itex]\cup[/itex](b,c) so infA = -∞ and supA = c. How did he find that A = (-∞,a)[itex]\cup[/itex](b,c) ?
The easiest way to solve this is by making a sign diagram (or whatever you call it).
Let me find out when (x-a)(x-b)<0.

If x<a, then x-a<0. If x>a, then x-a>0. If x<b, then x-b<0, If x>b, then x-b>0. So putting these in a diagram yields

------------------ a +++++++++++++++++++++++
-------------------------- b +++++++++++++++++

Multiplying the two gives us

++++++++++++++ a ------ b +++++++++++++++++

So we see that the function is negative between a and b. So A=(a,b) here.

Now try to find it for three terms.




2) B = {1 +(-1)n:n[itex]\in[/itex] N}. Find the supB and infB.

In the solution he says. Obviously B={0,1} which we can compute if we make n=2k (even) and n=2k+1 (odd)
so infB=0 and supB=1. So if n=2k wouldn't B be 2? Is this a mistake my professor made?
Yes, it needs to be B={0,2}
 
  • #3
Thanks for clearing these things up.
 

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