Q1. Let Show that A has a lower bound in but no greatest lower bound in . Give all details of the proof along the lines of the proof given in the lecture that the rationals are not complete.

Proof: Let . We aim to show that has a lower bound in but no greatest lower bound in .

Step 1: Existence of a Lower Bound in

Since all elements of are positive and satisfy , we observe that is a lower bound. For all , , so . Therefore, is a lower bound of in .

Step 2: Assume the Existence of a Greatest Lower Bound

To reach a contradiction, assume that there exists a rational number (where ) that is the greatest lower bound (infimum) of . We will examine two cases based on whether or .

Step 3: Case 1:

Suppose . We will show this leads to a contradiction by finding a rational number larger than that is still a lower bound of .

Since , there exists a positive integer such that:

Dividing both sides by , we obtain:

This implies that the rational number (which is larger than ) satisfies , making it a valid lower bound of .

Therefore, is larger than but still a lower bound of , which contradicts the assumption that is the greatest lower bound.

Step 4: Case 2:

Suppose , which implies (since and ).

Since , there exists an integer large enough such that:

This implies that , a rational number smaller than , is still in , as . This contradicts the assumption that is the greatest lower bound, as there is a smaller element in .

Step 5: Conclusion

Since both cases lead to contradictions, we conclude that there cannot be a greatest lower bound for in . Thus, has a lower bound in , but no greatest lower bound.

This completes the proof.

Let with . We need to show that there exists a rational number such that .

Q2. In addition to the completeness property, the Archimedean property is an important fundamentalproperty of R. It says is that if x, y ∈ and x, y > 0, there is an n ∈ such that nx > y.Use the Archimedean property to show that if r, s ∈ and r < s, there is a q ∈ such that r < q < s.(Hint: pick n ∈ N , , and find an m ∈ N such that .)

Proof: Step 1: Choose such that .

Since , by the Archimedean property, there exists an integer such that:

Step 2: Conclude that .

Since , we have . Adding to both sides gives:

Therefore, it suffices to find a rational number between and .

Step 3: Find an Integer such that .

Let be the greatest integer less than . Then we have:

Dividing by gives:

Step 4: Conclude that .

Since , it follows that:

Thus, is a rational number such that .

Conclusion: We have shown that for any with , there exists a rational number such that . This completes the proof. Q.E.D.

Q3. Formulate both in symbols and in words what it means to say that doesn’t approach as .

In Symbols

In Words

This expression states that:

There exists a positive number ( \epsilon ) such that no matter how large we choose ( m ), there will always be some ( n > m ) for which ( a_n ) is at least ( \epsilon ) away from ( a ).

In other words, the sequence ( a_n ) does not settle close to ( a ) as ( n ) grows, and there are always terms in the sequence that remain a fixed distance from ( a ), regardless of how far out in the sequence we go.

Q4. Prove that as .

Proof: To prove that as , let be given. We need to find an such that for all , .

Pick an so large that , by the Archimedean property. Then we have:
Now observe that:

Since , we have:

Therefore, for all , . This completes the proof.

Q5. Prove that as .

Proof: To prove that as , let be given. We need to find an such that for all , .

Choose so large that , by the Archimedean property. Then we have:
Now, for any , we observe that:

Therefore, for all , . This completes the proof.

Q6. Prove that as .

Proof: To prove that as , let be given. We need to find an such that for all , .

Choose so large that . This can be rearranged as:

Now, for any , we have:

Thus, we conclude that for all , .

This completes the proof.

Q7. Let be an increasing sequence (i.e., for each ). Suppose that as . Prove that .

Proof:Part 1: Prove That Is an Upper Bound of the Sequence
Goal: We need to prove that for all . This shows that is an upper bound of the sequence.
Proof by Contradiction: Assume there exists some such that:
Define:
Mathematical Induction: We need to prove that for any :
Base Case (): Since the sequence is increasing:
Subtracting from both sides:
Inductive Step: Assume that for some : Since the sequence is increasing:
Thus:
By induction, for all :
This implies that never stabilizes below or equal to if , which contradicts the convergence of the sequence to . Therefore:
Thus, is an upper bound of the sequence .

Part 2: Prove That Is the Least Upper Bound (Lub) by Contradiction

Goal: Prove that , where lub represents the least upper bound of the sequence.
Proof by Contradiction:Suppose there exists some that is an upper bound for the sequence .
Since , we can define:
Since as , for sufficiently large , we have:
This implies:
But since :
This implies that, for sufficiently large , . This contradicts the assumption that is an upper bound for the sequence .
No number can be an upper bound of the sequence .
Thus, is the least upper bound (lub) of the sequence.

Q8. Prove that if is increasing and bounded above, then it tends to a limit.

Proof: Bounded Above Implies Existence of Least Upper Bound

Let be an increasing sequence that is bounded above.
By the completeness property of real numbers, the set has a least upper bound , such that:
For every , there exists an index such that:

Convergence of to :Since is an increasing sequence, for any , we have:
Therefore,
Given that , it follows that:
Thus, for :
Conclusion:For any , there exists such that for all :
Hence, converges to .