Maximums and Minimums: Why is the Lower Limit Not 1 in This Set?

[hitmeoff]

Hello all,

I am currently taking Intro to Real Analysis and we are using Elementary Analysis by Ross. We are on Section 4, dealing with the completeness axiom, but we have only got as far as defining the minimum and maximum of a set (we have not discussed the completeness axiom, archimedean property, upper/lower bounds, supremums/infimums).

We went over Ross' examples in class and I have revied the examples myself and I am confused why the following is so:

1) The set {r in Q: 0 =< r =< Sqrt(2)} has a minimum, namely 0, but no maximum. This is because Sqrt(2) does not belong to the set, but there are rationals in the set arbitrarily close to Sqrt(2).

Ok, I get that. But the next example says:

2) Consider the set {n^(-1^n) : n in N}. This is shorthand for the set {1, 2, 1/3, 4, 1/5, 6, 1/7...} The set has no maximum and no minimum.

Ok, I understand why there is no maximum to the set, there are infinitely many more naturals that this set generates (ie. 8, 10, 12, 14.. -> infinity). What I do not understand is why is the lower limit not 1? 1 is the smallest possible natural number in this set. I know the set continues towards zero getting closer and closer to zero without ever reaching there, but those fractions are not naturals.

If in the first exampl2 Sqrt(2) is not the max because it is not a rational and you can get infinitely many rationals on your way to Sqrt(2) (ie. the max is restricted to rationals) why is the minimum of the second example not 1 (if we are restricting the min to naturals)?

Is it because the set in the first example is explicitly bounded by 1 and Sqrt(2) but the set in the second example is not explicitly limited?

 

[hitmeoff]

Anyone?

 

[HallsofIvy]

Hello all,

I am currently taking Intro to Real Analysis and we are using Elementary Analysis by Ross. We are on Section 4, dealing with the completeness axiom, but we have only got as far as defining the minimum and maximum of a set (we have not discussed the completeness axiom, archimedean property, upper/lower bounds, supremums/infimums).

We went over Ross' examples in class and I have revied the examples myself and I am confused why the following is so:

1) The set {r in Q: 0 =< r =< Sqrt(2)} has a minimum, namely 0, but no maximum. This is because Sqrt(2) does not belong to the set, but there are rationals in the set arbitrarily close to Sqrt(2).

Ok, I get that. But the next example says:

2) Consider the set {n^(-1^n) : n in N}. This is shorthand for the set {1, 2, 1/3, 4, 1/5, 6, 1/7...} The set has no maximum and no minimum.

Ok, I understand why there is no maximum to the set, there are infinitely many more naturals that this set generates (ie. 8, 10, 12, 14.. -> infinity). What I do not understand is why is the lower limit not 1? 1 is the smallest possible natural number in this set. I know the set continues towards zero getting closer and closer to zero without ever reaching there, but those fractions are not naturals.

It said "n in N". That does NOT mean that the numbers in the set are in N! The numbers in the set are, as it said, 1, 2, 1/3, etc. It would not say those numbers are in the set if it were intended to restrict the set to the integers.

If in the first exampl2 Sqrt(2) is not the max because it is not a rational and you can get infinitely many rationals on your way to Sqrt(2) (ie. the max is restricted to rationals) why is the minimum of the second example not 1 (if we are restricting the min to naturals)?

Because the first example specifically said that the numbers in the set were natural numbers. The second example said the number, n, used to generate the numbers in the set was restricted to the natural numbers but did NOT say that the numbers in the set had to be natural numbers and the numbers shown as belonging to the set are clearly NOT natural numbers.

Is it because the set in the first example is explicitly bounded by 1 and Sqrt(2) but the set in the second example is not explicitly limited?

No, the first set is clearly has 0 as a lower bound, whether "explicite" or not.

 

[hitmeoff]

Ok, I think I get it. Maybe I am getting confused on reading set notation.

So if we say the set S = {r in Q: 0 =< r =< Sqrt(2)} is like saying the set of rationals ranging from [0, Sqrt(2)] In which case 0 is a rational and the smallest value of the set so its the minumum and there is no maximum because there are infinitely many rationals on the way to Sqrt(2).

and

set T = {n^(-1^n) : n in N} is like saying the set of values generated by the function n^(-1^n) with the naturals as input. There is no min or max because there is no minimum or maximum value to the function.

Am I right on this? If I am, then let me point out a third example given by the professor:

{n in Z: 1 =< ln n =< 10} So do I read this "the set of values given by the function ln n, in the range [1, 10], with integers as the input"? So why is the min of this set 3? I know e^1 is 2.7... so the nearest integer is 3, but why are we restricting the min to just integers? obviously n can only be those numbers in the interval [e^1, e^10], I just don't get how 3 is even a member of this set? Is this set not: {ln 3, ln 4, ln 5...ln 22026} (e^10 = 22026.46...}

Or should I read {n in Z: 1 =< ln n =< 10} like "the set of integers n, subject to the function ln n in the interval [1,10]" in which case the integer 3 is the smallest integer that, when subject to ln n, gives a value that is in the interval [1,10]? Would the max of this set then be 22026?

 

[hitmeoff]

Or should I read {n in Z: 1 =< ln n =< 10} like "the set of integers n, subject to the function ln n in the interval [1,10]" in which case the integer 3 is the smallest integer that, when subject to ln n, gives a value that is in the interval [1,10]? Would the max of this set then be 22026?

In other words, is this set : {3, 4, 5,...22026}?