Homework 5

Solution.

Let $\p{s_n}$ be a bounded, decreasing sequence. Then the set $\set{s_n \mid n \in \N}$ is non-empty and bounded below, so $s = \inf S$ exists. We will show that $s_n \to s$.

Let $\varepsilon > 0$. Then $s + \varepsilon$ is not a lower bound for $\p{s_n}$, so there exists $N \in \N$ such that $s_N < s + \varepsilon$. Since $\p{s_n}$ is decreasing, we get

Moreover, $s_n - s \geq 0$ by definition of $s$, so for all $n \geq N$,

Solution.

I will give two solutions--one that just uses a theorem, and one that's more explicit in the construction. Either is fine, of course, but I suggest you understand the explicit construction.

Solution 1 (by Theorem 11.2)

$\Q$ is unbounded (e.g., $\N \subseteq \Q$), so because $\p{r_n}$ is an enumeration of $\Q$, it follows that $\p{r_n}$ itself is unbounded. Thus, by THeorem 11.2(ii), there exists a subsequence which diverges to $+\infty$.

Solution 2 (by construction)

Since $\p{r_n}$ is unbounded, $1$ is not an upper bound for $\p{r_n}$. Thus, there exists $n_1 \in \N$ such that $r_{n_1} \geq 1$.

We proceed by induction. Suppose we have chosen $n_1 < n_2 < \cdots < n_k$ such that $r_{n_k} \geq k$. Notice that

is still an unbounded set since we only remove finitely many elements. Thus, in particular, $k + 1$ is not an upper bound for $S_k$, so there exists $r_{n_{k+1}} \in S_k$ such that $r_{n_{k+1}} \geq k + 1$. By definition of $S_k$, we immediately have $n_{k+1} > n_k$ as well.

Thus, we have shown that there exists a subsequence $\p{r_{n_k}}$ such that $r_{n_k} \geq k$, which implies $r_{n_k} \to +\infty$.

Solution.

1. Let $\p{s_n}$ be a convergent sequence of elements in $\br{a, b}$ and let $s = \lim_{n\to\infty} s_n$. By definition, $a \leq s_n \leq b$, so by the lemma from discussion, $a \leq s \leq b$, which means $s \in \br{a, b}$.

2. No. Suppose there were such a sequence $\p{s_n}$. Recall that

   $$\limsup_{n\to\infty} s_n = \sup{\p{0, 1}} = 1.$$

   But we also showed that $\limsup_{n\to\infty} s_n$ itself is a subsequential limit, which means $1 \in \p{0, 1}$, which is impossible.