What does R\X mean?
Also, R is the set of all real numbers, right?
Attention Nerds
Do my homework for me please?
Exercise 1 (Due Monday, Oct. 9). Suppose X is a subset of R such that for the complement C(X) = R\X, neither X nor C(X) = the empty set; and the following is true:
(1) X U C(X) = R, and
(2) For all x in X and y in C(X), we have x < y.
Show carefully that there exists an a in R such that either
X = (−1, a] = {x in R| x less than or equal to a}
or
X = (−1, a) = {x in R| x < a}.
Some vague hints: What will be true about the a? The conclusion has cases, how will that affect the structure of the proof? What kind of statement are you trying to prove in the each case, that a set has a certain property? That two sets are equal? What is required to do that?
I'm sick of school. The assholes in the front office decided my hardest class needed to meet on Fridays at 9AM, when we all know Thursday night has the best drink specials for college students.
Jinheim
19 years ago
Bendir
19 years ago
R is the set of real numbers
R\X is the same as R - X, aka the complement of set X
R\X is the same as R - X, aka the complement of set X
Verileah
19 years ago
I just straight up don't remember enough math to be of use here, though I think there was a time I might have at the very least understood the question :(. I just wanted to say that 'Show carefully that there exists an a in R such that either...' amused me. What happens if you show carelessly? Does something explode?
Rikr
19 years ago
I'll take the zero. :(
Bendir
19 years ago
Well a is the supremum for X, that I know (because that's the subject of this chapter)
So here's my thinking for the two cases:
1) a is in X, so show a is an upper bound
2) a is in C(X) so show a is least upper bound
For (1) I want to assume for all x in X, show x
For (2) Proof by contrapositive, choose an arbitrary b in X and b
Can anyone tell me if there are any holes, have I missed something? If you feel like actually writing the proofs for the two cases, that'd be awesome too.
I'll probably just have to sit down and spend the required 30 minutes writing a proof and scrutinizing each step. This thread is all about my laziness and hatred for all things math. (I hope there's at least one math major on this forum)
So here's my thinking for the two cases:
1) a is in X, so show a is an upper bound
2) a is in C(X) so show a is least upper bound
For (1) I want to assume for all x in X, show x
For (2) Proof by contrapositive, choose an arbitrary b in X and b
Can anyone tell me if there are any holes, have I missed something? If you feel like actually writing the proofs for the two cases, that'd be awesome too.
I'll probably just have to sit down and spend the required 30 minutes writing a proof and scrutinizing each step. This thread is all about my laziness and hatred for all things math. (I hope there's at least one math major on this forum)
Onimi
19 years ago
math > me =(
Den
19 years ago
basic math is a piece of cake...algebra and I have NEVER gotten along...I'd even take geometry over algebra - blech.
Mirabela
19 years ago
set theory is great!
of course a exists in R
a is a member of x of X
using the trasitive property the set of X is a memeber of the set of R
that is of course a is not a complex number
but that is my primitive philosopher answer to that im sure they want a more mathematical one
of course a exists in R
a is a member of x of X
using the trasitive property the set of X is a memeber of the set of R
that is of course a is not a complex number
but that is my primitive philosopher answer to that im sure they want a more mathematical one
Xandare
19 years ago
i like...boobies