Proofs of the four type of intervals

[Kernul]

Homework Statement

Verify that if an interval is bounded it must be one of the four following types: $(a, b)$, $[a, b]$, $(a, b]$, or $[a, b)$.

Homework Equations

The Attempt at a Solution

I don't quite get what I should actually prove here.
Do I have to see if, for $A \subseteq S$, we have $\forall x \in S, a < x < b$, with $a$ and $b$ being the lower and upper bound of $A$? (this is for the first type $(a, b)$)

 

[Mark44]

Homework Statement

Verify that if an interval is bounded it must be one of the four following types: $(a, b)$, $[a, b]$, $(a, b]$, or $[a, b)$.

Homework Equations

The Attempt at a Solution

I don't quite get what I should actually prove here.
Do I have to see if, for $A \subseteq S$, we have $\forall x \in S, a < x < b$, with $a$ and $b$ being the lower and upper bound of $A$? (this is for the first type $(a, b)$)

How does your book define "bounded interval"?

For an interval of finite length, each endpoint either is or is not included in the interval.

 

[Kernul]

How does your book define "bounded interval"?

For an interval of finite length, each endpoint either is or is not included in the interval.

The book says "A given subset of R, and in particular an interval, is called bounded if it has both an upper and lower bound."
So I have to see if the endpoints $a$ and $b$ are the lower and the upper bound of the interval?

 

[Mark44]

The book says "A given subset of R, and in particular an interval, is called bounded if it has both an upper and lower bound."
So I have to see if the endpoints $a$ and $b$ are the lower and the upper bound of the interval?

You can assume that a is the lower bound and b is the upper bound. Now, is a included in the interval or is it not included? Is b included in the interval or is it not included? How many different interval types are we talking about here?

 

[fresh_42]

What is an interval?

 

[Kernul]

You can assume that a is the lower bound and b is the upper bound. Now, is a included in the interval or is it not included? Is b included in the interval or is it not included? How many different interval types are we talking about here?

Oh, so I start by assuming they are already a lower and upper bound and then prove if they are included in the interval or not. This means that I have to prove if the inequalities are $<$ or $\leq$.
The types are 4, right? Unless we just take the first and the third, since the other two are basically the opposite of the others.

What is an interval?

An interval is a subset $I$ of $\mathbb{R}$ where whenever $a$ and $b$ belong to $I$ and $a \leq c \leq b$, then $c$ also belongs to $I$.

 

[fresh_42]

I would start with a given set $S$ that is bounded, i.e. there are numbers $A$ and $B$ such that $A \leq s \leq B$ for all $s\in S$.
Next I would define supremum and infimum of these bounds, say $B_S$ and $A_I$, and proceed by what @Mark44 has said: they are either in or out of $S$. Finally there should be an argument, why there cannot be holes between $B_S$ and $A_I$, i.e. elements outside of $S$.

 

[Mark44]

I'm not sure that a full-fledged proof is required here.

Verify that if an interval is bounded...

Because of this wording, I believe "verify that ... " does not require the same rigor as "prove that ..." would need.

Next I would define supremum and infimum of these bounds

Posting in the Precalc section, the OP might not be familiar with these terms.

 

[Kernul]

I would start with a given set $S$ that is bounded, i.e. there are numbers $A$ and $B$ such that $A \leq s \leq B$ for all $s\in S$.
Next I would define supremum and infimum of these bounds, say $B_S$ and $A_I$, and proceed by what @Mark44 has said: they are either in or out of $S$. Finally there should be an argument, why there cannot be holes between $B_S$ and $A_I$, i.e. elements outside of $S$.

I get it. So I would find out if $B_S$ and $A_I$ are in or out of this set $S$.
The argument you're talking about is the axiom of completeness, right?

I'm not sure that a full-fledged proof is required here.

Because of this wording, I believe "verify that ... " does not require the same rigor as "prove that ..." would need.
Posting in the Precalc section, the OP might not be familiar with these terms.

So, what should I do when it says "verify that"? Simply see if they are in the interval or not?
Supremum and infimum are, respectively, the least upper bound and the greatest lower bound, right?

 

[fresh_42]

I get it. So I would find out if $B_S$ and $A_I$ are in or out of this set $S$.
The argument you're talking about is the axiom of completeness, right?

I'm not sure, if we really needed completeness, because I think, it should work for rational numbers as well: the left and right points of an interval are either in our out. But I might have overlooked a loophole for there are some really nasty sets out there. $B_S$ is simply the lowest upper bound of all possible and $A_I$ the largest lower bound of all possible. Completeness makes it easy, yes.

 

[Mark44]

I get it. So I would find out if $B_S$ and $A_I$ are in or out of this set $S$.
The argument you're talking about is the axiom of completeness, right?

IMO, not needed for this problem.

So, what should I do when it says "verify that"? Simply see if they are in the interval or not?

"they" being the endpoints -- how many ways can you specify an interval of finite length (i.e., a bounded interval) where the two endpoints either are or are not include? I believe that's all this problem is asking for.

Supremum and infimum are, respectively, the least upper bound and the greatest lower bound, right?

 

[Kernul]

"they" being the endpoints -- how many ways can you specify an interval of finite length (i.e., a bounded interval) where the two endpoints either are or are not include? I believe that's all this problem is asking for.

By saying that the endpoints are equal or greater than the lower bound and equal or lower than the upper bound.
Let's say $a, b$ are the endpoints. The least upper bound is $L$ and the greatest lower bound is $l$. Then I should have $a \geq l$ and $b \leq L$. Am I right?

 

[Mark44]

By saying that the endpoints are equal or greater than the lower bound and equal or lower than the upper bound.
Let's say $a, b$ are the endpoints. The least upper bound is $L$ and the greatest lower bound is $l$. Then I should have $a \geq l$ and $b \leq L$. Am I right?

No, that's not at all what I am saying, which has nothing to do with least upper bounds or greatest lower bounds. If you have an interval, is the left endpoint included? Yes or no?
Is the right endpoint included? Yes or no?
The yes/no answers determine four kinds of intervals.

 

[Kernul]

No, that's not at all what I am saying, which has nothing to do with least upper bounds or greatest lower bounds. If you have an interval, is the left endpoint included? Yes or no?
Is the right endpoint included? Yes or no?
The yes/no answers determine four kinds of intervals.

Oh. So it's really just that.
Sorry, I think I was trying to do something so simple in a complicated way.
Thank you.