I have this homework problem: Let $A \subseteq \mathbb{R}$ be a set of real numbers which is bounded above, and let $a = \sup A$. Show that there is a sequence $(a_n)$ so that $a_n \in A$ for each $n$ and $a_n \rightarrow a$. [Hint: Use the fact that $a − \frac{1}{n}$ is not an upper bound of $A$ to find $a_n$ , and check that the sequence you get converges to $a$.]

I understand the question what I don't get is how am I to properly use $a-\frac{1}{n}$ to determine $a_n$? I worked that through to arrive at $\frac{an-1}{n}$ but this seems to produce lunacy. For example,

$$ \begin{align} a_1 &= a-1 \\ a_2 &= \frac{a2-1}{2} \\ a_3 &= \frac{a3-1}{3} \\ \end{align} $$

Which is just gibberish. I've stared at this long enough; it's time for help. What simple thing am I not seeing?

  • 2
    $\begingroup$ Your original sequence might not include the value $a-1$, but it is certain to have an entry which is larger than $a-1$. You can pick that to be your $a_{1}$. $\endgroup$ – Brian Borchers Feb 8 '17 at 5:39
  • $\begingroup$ Hint: Use the fact that $a-\frac{1}{n}<a$ for all $n\in\mathbb{N}$, i.e., $a-1<a$, $a-\frac{1}{2}<a$, $a-\frac{1}{3}<a$. Then, you can construct a sequence using this fact. If $a-1\notin A$, and $a-\frac{1}{k}\notin A$, you can take the longest $n\in\mathbb{N}$ such that $a-\frac{1}{n}\in A$, then, for all $N\geq n$, holds $a-\frac{1}{N}\in A$ (via induction can be proved). The first $n$ therms in the sequence can be any element of $A$. $\endgroup$ – Carlos Jiménez Feb 8 '17 at 5:42

Using the hint, as $a$ is $\sup A$, there is at least one element in $A$ that is greater than $a-1$. If there weren't, $a-1$ would be an upper bound for $A$ that is less than $a$, contradicting the statement that $a=\sup A$. Pick any element greater than $a-1$ for $a_1$. Similarly, there is at least one element greater than $a-\frac 12$, so pick one for $a_2$. Each $a_n$ is chosen to be an element of $A$ that is greater than $a-\frac 1n$. Now you can do an $\epsilon-N$ proof that the sequence converges to $a$. If I give you $\epsilon$ you can say that all $a_k$ with $k \gt \frac 1\epsilon$ are within $\epsilon$ of $a$.

  • $\begingroup$ So, I took the implications of the hint to mean I had to find an expression for $(a_n)$ by using $a-\frac{1}{n}$. But what you're saying is that, because $a=\sup A$, then $\forall n\in\mathbb{N} : a-\frac{1}{n}$ is an element of $(a_n)$. Then, because $n\rightarrow\infty : \frac{1}{n}\rightarrow 0$ and thus the sequence converges to 0. $\endgroup$ – Andrew Falanga Feb 8 '17 at 15:27
  • $\begingroup$ No, you don't need to find an expression for $a_n$, you just need to show that an element exists in the range $(a-\frac 1n,a]$, then pick it for $a_n$. We don't know that $a-\frac 1n$ is an element of $A$. For example, $A$ could be the irrationals in $(0,1)$ with a $\sup$ of $1$. When we look for $a_3$ it is not true that $1-\frac 13 \in A$, but there is at least one irrational in $(\frac 23,1]$. We just say pick any one of them and make it $a_3$. Regardless of which one we pick, the sequence will converge to $1$. $\endgroup$ – Ross Millikan Feb 8 '17 at 15:47
  • $\begingroup$ In my previous comment, I didn't mean that $(a_n)$ converges to $0$ but that $(a_n)$ converges to $a$. $\endgroup$ – Andrew Falanga Feb 8 '17 at 17:46
  • 1
    $\begingroup$ Yes, that is right. We could even move away from $a$, as we might choose $a_1$ to be near $a-\frac 19$ then choose $a_2$ to be barely bigger than $a-\frac 12$, but eventually we get squeezed very close to $a$. It is important to note that we were not asked to construct a sequence, just to prove it exists. That is why we can say there is an element in some range, so pick one. We don't care which one is picked, we still come up with a sequence that converges to $a$. $\endgroup$ – Ross Millikan Feb 8 '17 at 17:53

All you need is the definition: $a = \sup(A)$ is the least upper bound of $A$.

This means that if somebody were to show you an upperbound $b$ of $A$ (so $\forall x \in A: x \le b)$, then you know for sure that $a \le b$.

This implies by taking the contrapositive: any number $b < a$ cannot be an upperbound for $A$! (or it would contradict the above)

Now let $a - \frac{1}{n} <a$ (for any $n$ we substract a strictly positive number). This is not an upperbound for $A$, so there is some $a_n \in A$ such that $a_n > a - \frac{1}{n}$ (or this smaller number would be an upperbound for $A$, which cannot be). You do know that $a_n \le a$, by virtue of $\sup(A)$ being an upperbound for $A$.

So you find a squeezed sequence $a_n \in (a - \frac{1}{n}, a]$ for every $n$. Now show $(a_n)_n$ must converge to $a$ (applying the definition of convergence, and using that for every $\varepsilon >0 ,\exists n: \frac{1}{n} < \varepsilon$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.