Advanced calculus proving assignment deadline tomorrow :(

[JanelSalas]

Please make your proofs concise.
1. If a<b, then a< (a+b)/2 < b
2. If I and J are open intervals, prove that either

I∩ J =Ø orI∩ J​

is also an open interval
3. Let x₁, x₂

∈​

. If x₁ is not equal to x₂, then there exists ε₁>0 and ε₂>0 such that N(x₁, ε₁)

∩​

N( x₂, ε₂) =

Ø​

4. Let β be a lower bound for E

⊂​

. Then β = inf E if and only if for every ε>0, there exists x

∈​

E such that x <β +ε
5. Let E be a bounded set. If F

⊂​

E, then F is also bounded. Furthermore, sup F

≤​

sup E and inf E

≤​

inf F

PLEASE HELP ME TO PROVE THIS. :?

 

[stapel]

Please make your proofs concise.

Um... Doing the actual exercises is actually your job. We're just here to help (like you saw in the "Read Before Posting" thread that you read before posting).

\(\displaystyle \mbox{1. If }\, a\, <\, b,\, \mbox{ then }\, a\, <\, \dfrac{a\, +\, b}{2}\, <\, b.\)

The middle term in the compound inequality is the average of the two variables' values. What can you say about the "average" of "a" and "a"? Of "b" and "b"? How do these "averages" relate to the end terms of the inequality? To the middle term?

\(\displaystyle \mbox{2. If }\, I\, \mbox{ and }\, J\, \mbox{ are open intervals, prove that either }\, \)\(\displaystyle I\, \cap\, J\, =\, \emptyset\, \mbox{ or else that }\, I\, \cap\, J\, \)\(\displaystyle \mbox{ is also an open interval.}\)

I think a proof by contradiction might be helpful here. If the intersection is not empty, then assume that the intersection is closed. What then can you conclude?

\(\displaystyle \mbox{3. Let }\, x_1, x_2\, \in\, \mathbb{R}.\, \mbox{ If }\, x_1\, \neq\, x_2,\, \mbox{ then prove that there exist }\, \epsilon_1\, >\, 0\, \mbox{ and }\, \epsilon_2\, \)\(\displaystyle \mbox{ such that }\, N(x_1,\, \epsilon_1)\, \cap\, N(x_2,\, \epsilon_2)\, =\, \emptyset .\)

Would it be correct to assume that "N(a, b)" means "the neighborhood, centered at 'a' with radius 'b'"? If so, what did you get when you assumed that the intersection was not empty, and then chased the elements of the intersection?

\(\displaystyle \mbox{4. Let }\, \beta\, \mbox{ be a lower bound for }\, \)\(\displaystyle E\, \subset\, \mathbb{R}.\, \mbox{ Then prove that } \beta\, =\, \mbox{ inf }\, E\, \Longleftrightarrow \, \forall\, \epsilon\, >\, 0,\, \exists\, x\, \in\, E\, \mbox{ such that }\, x\, <\, \beta\, +\, \epsilon.\)

What is your book's definition of the lower bound of a set? What is its definition of "infimum"? What have you gotten, in either direction, based on these definitions?

\(\displaystyle \mbox{5. Let }\, E\, \mbox{ be a bounded set. If }\, F\, \subset\, E,\, \mbox{ prove that }\, F\, \)\(\displaystyle \mbox{ is also bounded. Prove furthermore that sup }\, F\, \leq\, \mbox{ sup }\, E\, \mbox{ and that inf }\, E\, \leq\, \mbox{ inf }\, F.\)

What is your book's definition of a bounded set? How far have you gotten in the first part of this proof?

Please be complete, so we can see where you're getting into trouble. Thank you!

 

[pka]

Please make your proofs concise.
1. If a<b, then a< (a+b)/2 < b
2. If I and J are open intervals, prove that either I∩ J =Ø orI∩ J is also an open interval
3. Let x1, x2∈R, then there exists ε1>0 and ε2>0 such that N(x1, ε1)∩N( x2, ε2) =Ø
5. Let E be a bounded set. If F⊂ E, then F is also bounded. Furthermore, sup F≤sup E and inf E≤inf F

1)
\(\displaystyle a<b \Rightarrow\left\{\begin{align*} \frac{a}{2}&<\frac{b}{2}\\\frac{a}{2}+\frac{a}{2}&<\frac{b}{2}+\frac{a}{2}\\ a&\le \dfrac{a+b}{2}\end{align*}\right. \) Now complete the proof.

While I do agree that you must follow the definitions given, there are traditional ones that simplify proofs.
All of this is in \(\displaystyle \mathbb{R}^1\). The statement that \(\displaystyle I\) is an open interval means that there are two real numbers such that \(\displaystyle a<b\) & \(\displaystyle I=(a,b)=\{x:a<x<b\}\). So suppose that \(\displaystyle I=(a,b)~\&~J=(c,d)\).
2) If \(\displaystyle I\cap J\not=\emptyset\) then \(\displaystyle I\cap J=(\max\{a,c\},\min\{b,d\})\)

Using this definition makes the other proofs simple. In most cases, if this statement is not the definition then it is easily proven as a limma.