Proving inf(ST) = inf(S)*inf(T)

  • Thread starter faceblah
  • Start date
  • #1
5
0

Homework Statement



Sets A and B are sets of positive real numbers. Define C = {st| s \in S and t \in T}
Prove inf(C) = inf(S)*inf(T)

The Attempt at a Solution



so i'm trying to prove inf(C) <= inf(S)*inf(T) and inf(C) >= inf(S)*inf(T).

i'll use e as epsilon. epsilon is positive

By definition there is an s in S such that: s < inf S + e. There is a t in T such that: t < inf T + e. Additionally, inf(C) <= st for all s in S and t in T by definition.

st < (inf S + e)(inf T + e) = (inf S)(inf T) + (inf S)*e + (inf T)*e + e^2

inf C <= st <= (inf S)(inf T) < inf S)(inf T) + (inf S)*e + (inf T)*e + e^2
(NOTE: i'm not sure about the middle inequality: "st <= (inf S)(inf T)" )

is this a correct way to go about this? I'm also not sure about proving the other direction
 

Answers and Replies

  • #2
22,129
3,297

Homework Statement



Sets A and B are sets of positive real numbers. Define C = {st| s \in S and t \in T}
Prove inf(C) = inf(S)*inf(T)

The Attempt at a Solution



so i'm trying to prove inf(C) <= inf(S)*inf(T) and inf(C) >= inf(S)*inf(T).

i'll use e as epsilon. epsilon is positive

By definition there is an s in S such that: s < inf S + e. There is a t in T such that: t < inf T + e. Additionally, inf(C) <= st for all s in S and t in T by definition.

st < (inf S + e)(inf T + e) = (inf S)(inf T) + (inf S)*e + (inf T)*e + e^2

inf C <= st <= (inf S)(inf T) < inf S)(inf T) + (inf S)*e + (inf T)*e + e^2
(NOTE: i'm not sure about the middle inequality: "st <= (inf S)(inf T)" )

is this a correct way to go about this? I'm also not sure about proving the other direction

No, the middle inequality is indeed not correct. Try it like this:

[tex]inf(C)\leq st<inf(S)inf(T)+inf(S)e+inf(T)e+e^2[/tex]

Because e is arbitrary, we can let e go to 0, thus

[tex]inf(C)<inf(S)\inf(T)[/tex]

For the other inequality, take [itex]st<inf(C)+e[/itex] and do something with it.
 
  • #3
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,950
19
st < (inf S + e)(inf T + e) = (inf S)(inf T) + (inf S)*e + (inf T)*e + e^2

inf C <= st <= (inf S)(inf T) < inf S)(inf T) + (inf S)*e + (inf T)*e + e^2
(NOTE: i'm not sure about the middle inequality: "st <= (inf S)(inf T)" )
You're right to be unsure. You have
  • [itex]\inf C \leq st[/itex]
  • [itex]st < (\inf S)(\inf T) + (\inf S) \epsilon + (\inf T) \epsilon + \epsilon^2[/itex]
So you just put the two together:
[itex]\inf C \leq st< (\inf S)(\inf T) + (\inf S) \epsilon + (\inf T) \epsilon + \epsilon^2[/itex]​
well, what you really care about is just transitivity:
[itex]\inf C < (\inf S)(\inf T) + (\inf S) \epsilon + (\inf T) \epsilon + \epsilon^2[/itex]​
And then invoke what you can about the fact that this is true for every positive real number [itex]\epsilon[/itex].

(aside: micromass forgot that < turns into [itex]\leq[/itex] when doing limits)
 
  • #4
22,129
3,297
(aside: micromass forgot that < turns into [itex]\leq[/itex] when doing limits)

Oh my, I'm still not fully awake :frown:
 
  • #5
5
0
For the other direction I'm guessing it's:

There is an st in C such that st < inf C + e.
By definition, inf S <= s for all s and inf T <= t for all t. So (inf S)(inf T) <= st.

So we have (inf C + e) > st => (inf S)(inf T). so (inf C + e) > (inf S)(inf T). (I need this to be a => though)

I'm not quite sure what you mean by "invoking what you know about e". Does this mean that the ">" becomes a "=>". See below for context

well, what you really care about is just transitivity:
[itex]\inf C < (\inf S)(\inf T) + (\inf S) \epsilon + (\inf T) \epsilon + \epsilon^2[/itex]​
And then invoke what you can about the fact that this is true for every positive real number [itex]\epsilon[/itex].

(aside: micromass forgot that < turns into [itex]\leq[/itex] when doing limits)
 

Related Threads on Proving inf(ST) = inf(S)*inf(T)

  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
14
Views
3K
  • Last Post
Replies
0
Views
2K
Replies
1
Views
922
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
1
Views
992
Replies
4
Views
2K
  • Last Post
Replies
0
Views
2K
Top