- #1
hitmeoff
- 261
- 1
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?
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?