Set of real numbers, bounds and sequences II

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 In this post we carry on from where we got to in Part I. The aim in this part is to try to tie together what we know about infimum, supremum, epsilons and sequences. This will give us a lot of flexibility in how we approach problems later: we can choose which criterion will be easiest to work with for any given situation.

Although working with sequences instead of epsilon and delta can lead to some very clean proofs, you shouldn't forget about epsilon and delta! That approach is very important later throughout mathematics, and it is best to be fluent in all possible methods so that you have the most suitable method available when you need it.

Let's start by having another look at supremum. The results for infimum are almost identical: you just need to flip everything e.g. by reflecting across $0$ in the number line using the map $x \mapsto -x$, which reverses all of the relevant inequalities.

Let $A\subseteq \R$ be a non-empty set which is bounded above in $\R$, and set $S = \sup A \in \R$, the supremum or least upper bound for $A$. We don't know whether $S\in A$ or not. If $S\in A$ then $S$ is the maximum element of $A$. If $S \notin A$, then $A$ does not have a maximum element.

(In general, you can easily check that whenever $A$ does have a maximum element $a_\max \in A$, we always have $a_\max = \sup A \in A.$)

Returning to the situation above, we have $S=\sup A$, but we don't know yet whether $S \in A$ or not.

(Both can happen of course. You should look at all examples of sets you have seen before to see what happens. Some examples of non-empty sets which are bounded above that you could consider are the sets $(-\infty,5]$, $[2,7)$, $\{-1/n: n \in \N\}$, $\{(-1)^n/n: n \in \N\}$.)

Anyway, we have $S=\sup A$. You can now check easily that the set of all upper bounds for $A$ is precisely the set $B=[S,\infty)$. So the set $B$ of upper bounds for $A$ in $\R$ is bounded below by $S$, and the set $B$ does have a minimum element, because $S \in B$. So we have $S=\inf B = \min B$. 

Exercise: Sketch these sets $A$ and $B$ on a number line for some of the specific examples suggested above.

Of course this means that the interval $(-\infty,S)$ is precisely the set of those real numbers which are NOT upper bounds for $A$. So for all $y \in (-\infty,S)$, we have that $y$ is NOT an upper bound for $A$, and so, negating the definition of upper bound, there exists $a \in A$ with $a>y$.

Let's summarise the situation for a general point $y \in \R$. We have:

  • $y$ is an upper bound for $A$ if and only if, for all $a \in A$, we have $a\leq y$;
  • $y$ is not an upper bound for $A$ if and only if there exists $a\in A$ such that $a>y$;
  • $y$ is the least upper bound for $A$, $\sup A$, if and only if we have both that $y$ is an upper bound for $A$ and also, for all $y'<y$, $y'$ is not an upper bound for $A$.
So where do $\epsilon$ and sequences come into the picture? Well another way of thinking about a real number that is (slightly) smaller than $S$ is to call it $S-\epsilon$, where $\epsilon$ is a (usually small but) strictly positive real number. We only really care about the small $\epsilon$s, because for example if $a>S-1/10$ then we also have $a>S-1$: knowing that $S-\epsilon$ is not an upper bound for $A$ for small epsilons also takes care of the larger epsilons too!

In fact, it is enough to check the special cases where $\epsilon = 1/n$, for $n=1,2,3,\dots$ (or you can use any sequence of positive real numbers which tend to $0$), because again the small $\epsilon$s take care of the large ones.

With $A$, $B$ and $S$ as above and $n \in \N$, we know that $S-1/n$ is not an upper bound for $A$ (because $S$ is the least upper bound for $A$), so we can choose an $a_n \in A$ with $a_n>S-1/n$. However we also know that $a_n \leq S$ (because $S$ is an upper bound for $A$). So in fact we have

\[S-\frac1n < a_n \leq S\,,\]

and so it follows (e.g. by the Squeeze Rule/Squeeze Theorem/Sandwich Theorem, or directly from the definitions) that $a_n \to S$ as $n \to \infty$.

Conversely, let $y \in B$, i.e., $y \in \R$ and $y$ is an upper bound for $A$. Of course we have $y\geq S$. Suppose now that there exists a sequence $(a_n)$ of points of $A$ which converges to $y$. Since $a_n \leq S$ for all $n$, this is also true in the limit, and so we have $y \leq S$. Combined with the above, we obtain $y=S$. Our conclusion is the following very useful lemma which helps once you have already established that a real number $y$ is an upper bound for $A$.

Lemma 1 (characterisations of supremum among all upper bounds in terms of $\epsilon$ and in terms of sequences)

Let $A$ be a nonempty subset of $\R$ and suppose that $A$ is bounded above. Let $y\in \R$ be an upper bound for $A$. Then the following statements are equivalent.

(a) For all $\epsilon>0$, we have that $y-\epsilon$ is not an upper bound for $A$.

(b) For all $\epsilon>0$, there exists $a \in A$ with $a>y-\epsilon$.

(c) There exists a sequence $(a_n)$ of elements of $A$ which converges to $y$ in $\R$.

(d) We have $y=\sup A$, i.e., $y$ is the supremum of $A$.

As mentioned earlier, the situation for infimum is exactly the same but flipped. You can check for yourselves the following characterisations of infimum.

Lemma 2 (characterisations of infimum among all lower bounds in terms of $\epsilon$ and in terms of sequences)

Let $A$ be a nonempty subset of $\R$ and suppose that $A$ is bounded below. Let $y\in \R$ be a lower bound for $A$. Then the following statements are equivalent.

(a) For all $\epsilon>0$, we have that $y+\epsilon$ is not a lower bound for $A$.

(b) For all $\epsilon>0$, there exists $a \in A$ with $a< y+\epsilon$.

(c) There exists a sequence $(a_n)$ of elements of $A$ which converges to $y$ in $\R$.

(d) We have $y=\inf A$, i.e., $y$ is the infimum of $A$.

These two lemmas are very helpful when proving results in analysis (e.g. for continuous functions) using a combination of sequences and supremum/infimum.

Warning! Unless you have already established that $y$ is an upper bound for $A$, you can't apply Lemma 1. (Similarly for Lemma 2 and lower bounds.) For example consider one of the specific sets mentioned above, 

\[A=\left\{\frac{(-1)^n}n:n\in\N\,\right\}= \{-1,1/2,-1/3,1/4,\dots\,\}\,.\]

In this case the (perhaps) most obvious sequence of elements of $A$ converges to $0$. But $0$ is neither an upper bound nor a lower bound for $A$ here! In this particular example, the sequences of elements of $A$ that you need will have to have infinitely many repeated elements (and indeed will have to be 'eventually constant') in order to converge to either $\inf A$ or $\sup A$.

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