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5. Midterm 1 Review Solutions

≪ 4. Midterm 1 Review | Table of Contents | 6. \(\limsup\), \(\liminf\), and Cauchy sequences ≫

Here’s a set of solutions to the first 7 problems of the practise problems for the first midterm. I didn’t give the last problem in class, so I’ll leave it as an exercise if you’re interested.

For the example where they’re not equal, let \(S = \left\lbrace 1, 3 \right\rbrace\) and \(T = \left\lbrace 1, 2 \right\rbrace\). Then \(S\cap T = \left\lbrace 1 \right\rbrace\), so \(\sup \left( S\cap T \right) = 1\) while \(\min \left\lbrace \sup S, \sup T \right\rbrace = 2\).

To show the inequality, we have that for all \(x\in S\cap T\), \(x\in S\) and thus \(x \leq \sup S\). However we also have \(x\in T\), so \(x \leq \sup T\) (the supremum of a set is an upper bound). Thus, for all \(x\in S\cap T\), \(x \leq \min \left\lbrace \sup S, \sup T \right\rbrace\), and the right hand side is an upper bound of \(S\cap T\).

Since the supremum is the least upper bound, we conclude that \(\sup \left( S\cap T \right) \leq \min \left\lbrace \sup S, \sup T \right\rbrace\). \(\square\)

For any \(\epsilon > 0\), we need to find an \(N_ \epsilon \) such that for all \(n > N_ \epsilon \), \[\left\lvert \frac{an+c}{bn+d} - \frac{a}{b} \right\rvert < \epsilon . \] We can simplify the difference on the left side: \(\frac{a}{b} = \frac{an+\frac{ad}{b}}{bn+d}\), so \[\frac{an+c}{bn+d} - \frac{a}{b} = \frac{c - \frac{ad}{b}}{ bn+d }.\] For all \(n\) sufficiently large (i.e. \(n > N_ \epsilon = \frac{1}{\left\vert b\right\rvert \epsilon }\left\lvert c- \frac{ad}{b} \right\rvert + \left\lvert \frac{d}{b} \right\rvert \) explicitly, I think), this quantity will be smaller than \(\epsilon \). \(\square\)

Note that I wouldn’t expect you to work out the explicit value of \(N_ \epsilon \) on a problem like this in an exam environment. I’m happy if you just say “for all \(n\) sufficiently large…”

This one’s a bit tricky, and it’s critical that \(S\) and \(T\) don’t contain any negative numbers.

First, we have \(\sup(S)\sup(T)\) is an upper bound of \(S\times T\). If \(s t \in S \times T\) for some \(s\in S\) and \(t\in T\), we have \(s\leq \sup S\) and \(t \leq \sup T\). Thus \(s t \leq \sup(S) \sup (T)\), as claimed.

Next, we claim that for any \(\epsilon > 0\), \(\sup(S) \sup(T) - \epsilon \) is not an upper bound of \(S\times T\). Morally speaking, we should have something like \[\sup(S) \sup (T) - \epsilon \approx \left( \sup S - \epsilon’ \right) \left( \sup T - \epsilon’ \right),\] i.e. it’s the product of two dudes slightly smaller than \(\sup S\) and \(\sup T\). Indeed, we have \[\begin{align*} \left( \sup S - \epsilon ’ \right) \left( \sup T - \epsilon ’ \right) & = \sup(S) \sup (T) - \epsilon ’ \left( \sup S + \sup T \right) + \left( \epsilon ’ \right) ^2 \\ & \geq \sup(S) \sup (T) - \epsilon , \end{align*} \] where we pick \(\epsilon ’ = \frac{\epsilon }{\sup S + \sup T} > 0\).

There exist \(s\in S\) and \(t\in T\) such that \[\begin{align*} \sup S - \epsilon ’ < s \leq \sup S && \textrm{and} && \sup T - \epsilon ’ < t \leq \sup T.\end{align*} \] Then, we get \[s t > \left( \sup S - \epsilon ’ \right) \left( \sup T - \epsilon ’ \right) > \sup(S) \sup(T) - \epsilon.\] By definition, \(s t \in S \times T\). Since \(\sup(S) \sup(T) - \epsilon \) is not an upper bound for \(S\times T\) for any \(\epsilon > 0\), we conclude that \(\sup \left( S\times T \right) = \sup(S) \sup(T)\). \(\square\)

Take \(S = (0,1 )\) and \(T = \left\lbrace 1 \right\rbrace\).

The point of this exercise is to remind you that in general, the supremum of a set is not contained in that set!

Let \(x_n\) be a sequence of one \(1\), two \(0\)’s, four \(1\)’s, eight \(0\)’s, etc. Then \(S _{2^n}\) fluctuates between \(\approx \frac{1}{2}\) and \(\approx \frac{1}{4}\), thus never converges. (I’ll leave writing up the gritty details up to you.)

We’ll prove the “only if” with the contrapositive.

Let \(M > 0\). Suppose there is some \(\epsilon > 0\) such that there are only finitely many \(n\) for which \(x_n > M - \epsilon \). Then, there exists some \(N_ \epsilon \) such that for all \(n > N _ \epsilon \), \(x_n \leq M - \epsilon \). That is, \[\sup \left\lbrace x_n : n > N _ \epsilon \right\rbrace \leq M - \epsilon .\] However, since \[\limsup _{n\to\infty} x_n = \lim _{N} \left( \sup \left\lbrace x_n : n > N \right\rbrace \right), \] and since \[\sup \left\lbrace x_n : n > N \right\rbrace \leq \sup \left\lbrace x_n : n > N _ \epsilon \right\rbrace\] for all \(N > N_ \epsilon \), it follows that \(\limsup _{n\to\infty} x_n \leq M - \epsilon \). In particular, \(M \neq \limsup _{n\to\infty} x_n\).

For the “if” direction, suppose \(M\) satisfies the assumptions above. Let \(\epsilon > 0\). By assumption, there exists \(N_ \epsilon \) such that \(x_n < M + \epsilon \) for all \(n > N_ \epsilon \). Thus, \[\sup \left\lbrace x_n : n > N \right\rbrace \leq M + \epsilon \] for all \(N \geq N _ \epsilon \), hence \(\limsup _{n\to\infty} x_n \leq M + \epsilon \) for all \(\epsilon > 0\). It follows that \(\limsup _{n\to\infty} x_n \leq M\).

On the other hand, for any \(\epsilon > 0\), and any \(N\), there must exist \(n > N\) such that \(x_n > M - \epsilon \). Otherwise, \(x_n > M - \epsilon \) at most \(N\) times, and our assumption forbids this. That is to say, \[\sup \left\lbrace x_n : n > N \right\rbrace \geq M - \epsilon \] for all \(N\), thus \(\limsup _{n\to\infty} x_n \geq M - \epsilon \) for all \(\epsilon > 0\). We conclude that \(\limsup _{n\to\infty} x_n \geq M\), and together with the previous inequality, it follows that \(\limsup _{n\to\infty} x_n = M\). \(\square \)

First, we note that statements (1) and (2) imply (3) by the monotone convergence theorem.

It will turn out to be easier to prove (1) after proving (2), so we’ll start by showing the sequence is bounded.

Let \(\phi = \frac{1+ \sqrt 5}{2}\), the golden ratio (and in fact, the limit of this sequence, but we don’t know that yet). \(\phi \) satisfies the equation \(\phi ^2 - \phi - 1 = 0\), so \(\phi + 1 = \phi ^2\) (we’ll use this in a bit). We claim that \(x_n \leq \phi \) for all \(n\), and we’ll prove this by induction.

• Base case: for \(n = 1\), this is trivially true.
• Inductive step: suppose \(x _{n-1} \leq \phi \) for some \(n > 1\). Then, \(x _{n-1} + 1 \leq \phi + 1 = \phi ^2\), hence \(x _{n}^2 \leq \phi ^2 \implies x_n \leq \phi \).

By induction, we conclude that \(x_n \leq \phi \) for all \(n\).

For part (1), one can show that for all \(0 \leq x \leq \phi \), \(\sqrt{x + 1} \geq x\), e.g. by squaring both sides and doing the algebra (computation omitted for sanity). By part (2), we know \(x_n \leq \phi \) for all \(n\). Thus, \[x_n = \sqrt{x _{n-1} + 1} \geq x _{n-1}\] for all \(n\). \(\square\)

Sorry, it turns out this problem was a bit harder than I had anticipated. You need a really tight upper bound to make monotonicity work. I’m interested to hear if you have a better solution than me though!