Show directly that if $(s_n)$ is decreasing sequence, then $\lim s_n = \inf\{s_n : n \in \mathbb{N} \}$.

Question

The metric space is (X, d) where $X \neq \emptyset$ and all subsets of $\mathbb{R}^k$ have the Euclidean metric.

I was also told to consider the cases where $s_n$ is bounded below and another where it isn't.

Here is my take on the case where $s_n$ is bounded:

Assume $s_n$ is decreasing and bounded. By upper lower bound property, $\lim s_n = \inf\{s_n : n \in \mathbb{N} \}$ exists in $\mathbb{R}$. Let $\epsilon>0$, then $s+\epsilon$>s and so $s+\epsilon$ is not a lower bound. Hence, $\exists N \in \mathbb{N}$ such that $s_n > s - \epsilon$, but then $\forall n \geq N$, $s+\epsilon > S_n > S_N > S > S - \epsilon$, and so $|S_N - S| < \epsilon$. Since $\epsilon$ was abritrary, $ \lim s_n = \inf \{s_n : n \in \mathbb{N} \}$.

However, I'm having difficulty with the case where $s_n$ is not bounded. I know that if $s_n$ is unbounded and decreasing, then $(s_n)$ diverges to $-\infty$. However, I'm a little bit at a lost as to how to prove this.

Help would be much appreciated!

Answer 1

Your first sentence doesn’t make much sense: the problem is clearly about sequences in $\Bbb R$. I’d just delete it. Now let’s look at what you have for the bounded case.

Assume $s_n$ is decreasing and bounded. By upper lower bound property, $\lim s_n = \inf\{s_n : n \in \mathbb{N} \}$ exists in $\mathbb{R}$.

This isn’t quite right. You know that $\inf\{s_n:n\in\Bbb N\}\in\Bbb R$, but you don’t know that it’s equal to $\lim_ns_n$: that’s precisely what you’re trying to prove.

Let $\epsilon>0$, then $s+\epsilon>s$ and so $s+\epsilon$ is not a lower bound.

You haven’t told us what $s$ is. Presumably you want $s=\inf\{s_n:n\in\Bbb N\}$, but you need to say so.

Hence, $\exists N \in \mathbb{N}$ such that $s_n > s - \epsilon$,

This isn’t what you wanted to say here: you already know that $s_n\ge s>s-\epsilon$ for each $n\in\Bbb N$ and $\epsilon>0$. Moreover, there seems to be no relationship between the $N$ in your quantifier and the subscript in $n$ in $s_n>s-\epsilon$. Presumably you intended to say that there is an $N\in\Bbb N$ such that $s_N<s+\epsilon$.

but then $\forall n \geq N$, $s+\epsilon > S_n > S_N > S > S - \epsilon$,

This should be $s+\epsilon>s_N\ge s_n\ge s>s-\epsilon$. (It’s likely that decreasing means weakly decreasing, i.e., non-increasing.)

and so $|S_N - S| < \epsilon$.

You mean that $|s_N-s|<\epsilon$. But you knew that already, when you chose $N\in\Bbb N$ such that $s_N<s+\epsilon$: $s+\epsilon>s_N\ge s$, so $|s_N-s|=s_N-s<(s+\epsilon)-s=\epsilon$. This has nothing to do with the statement beginning ‘but then $\forall n\ge N$’, which serves no purpose here at all.

Since $\epsilon$ was arbitrary, $ \lim s_n = \inf \{s_n : n \in \mathbb{N} \}$.

Here is where the fact that $s+\epsilon>s_n>s-\epsilon$ for each $n\ge N$ is relevant.

Apart from some badly mangled notation you appear to have the basic pieces, but you need to put them together much more carefully, something like this:

Assume that $\langle s_n:n\in\Bbb N\rangle$ is decreasing and bounded. Then $\{s_n:\in\Bbb N\}$ has a lower bound and therefore has a greatest lower bound $s=\inf\{s_n:n\in\Bbb N\}\in\Bbb R$; we want to show that $s=\lim_ns_n$.

Let $\epsilon>0$. Then $s+\epsilon>s$ is not a lower bound for $\{s_n:n\in\Bbb N\}$, so there is an $N\in\Bbb N$ such that $s_N<s+\epsilon$. The sequence is decreasing, so $s-\epsilon<s\le s_n\le s_N<s+\epsilon$ for each $n\ge N$, and hence $|s_n-s|<\epsilon$ for each $n\ge N$. Since $\epsilon>0$ was arbitrary, it follows that $s=\lim_ns_n$.

Now we can deal with the unbounded case.

Since the sequence is decreasing, $s_n\le s_0$ for each $n\in\Bbb N$, and $s_0$ is therefore an upper bound (indeed, the least upper bound) for the set $\{s_n:n\in\Bbb N\}$. Thus, if this set is not bounded, that is because it has no lower bound in $\Bbb R$, in which case by definition $\inf\{s_n:n\in\Bbb N\}=-\infty$. That means that for each $x\in\Bbb R$ there is an $n_x\in\Bbb N$ such that $s_{n_x}<x$. But then $s_n\le s_{n_x}<x$ for each $n\ge n_x$, so by definition

$$\lim_ns_n=-\infty=\inf\{s_n:n\in\Bbb N\}\;.$$