I came across the following problems during my course of study in real analysis:

If $(A_n)$ is an increasing sequence of sets with union $A$ (i.e. $A_n \uparrow A$), then show that $A_n \to A$. State and prove the analogous proposition for decreasing sequences.

So $A_1 \subseteq A_2 \subseteq \cdots \subseteq A_n \subseteq \cdots$ and $A_1 \cup A_2 \cup \cdots \cup A_n \cup \cdots = A$. We want to show that $\text{lim inf} \ A_n = \text{lim sup} \ A_n = A$. Now if $x \in A_n$ for any $n$ then $x \in A$ by definition of union. Hence the result follows?

The analogous result would be: $A_n \downarrow A$ and $\bigcap A_n = A$, then $A_n \to A$? I suppose we could also use the intersection/union definition of limit supremums and limit infimums.

If $(a_n)$ and $(b_n)$ are null sequences, show that $(a_{n}b_{n})$ is a null sequence.

So for every $\epsilon >0$, $|a_n| \leq \epsilon$ for all $n > N_1$ and $|b_n| \leq \epsilon$ for all $n> N_2$. So choose $N_3 = \max \{N_1, N_2 \}$. Thus for every $\epsilon >0$, $|a_n b_{n}| \le \epsilon^{2} < \epsilon$ for all $n > N_3$?

Do these ideas sound correct?

  • 1
    $\begingroup$ Shouldn't you get $|a_nb_n|<\epsilon^2$ in the previous to last line? $\endgroup$ – Andrés E. Caicedo Jun 23 '11 at 5:35
  • $\begingroup$ @Andres Caicdeo: Which then implies that $|a_{n} b_{n}| < \epsilon$. $\endgroup$ – Damien Jun 23 '11 at 5:37
  • $\begingroup$ @Arturo: That's what I wrote first then I switched it. $\endgroup$ – Damien Jun 23 '11 at 5:40
  • $\begingroup$ @Damien: You've posted here two questions that have little or nothing to do with one another, except for the fact that it's you who are asking them (and the contain the word sequence). It's best if multiple questions in the same post are closely related to one another. $\endgroup$ – Arturo Magidin Jun 23 '11 at 5:46
  • $\begingroup$ For those who are unfamiliar with the notation "limsup" and "liminf" when applied in the context of sets: $\limsup A_n =\{x:x\in A_n \text{ for infinitely many }n\}$ and $\liminf A_n=\{x:\text{there exists N such that }x\in A_n\text{ for all }n\geq N\}$. $\endgroup$ – Amitesh Datta Jun 23 '11 at 5:49

Note. The question originally had $A=A_1\cup\cdots\cup A_n$, terminating at $n$.

You misintepreted the notation in the first problem. You have a sequence of sets: one for each natural number. So you h ave $$A_1\subseteq A_2\subseteq A_3\subseteq \cdots \subseteq A_n\subseteq A_{n+1}\subseteq \cdots$$ and $$A = \bigcup_{i=1}^{\infty} A_i = A_1\cup A_2\cup A_3\cup\cdots \cup A_n\cup A_{n+1}\cup\cdots$$ I don't know what your definition of a set converging to another set is, though.

Your guess as to what the corresponding result for decreasing intersections is correct (once you correct the possible misconception of exactly what it is that what you typed means).

As for the second problem, $|a_nb_n|\lt \epsilon^2$ only implies $|a_nb_n|\lt \epsilon$ if $\epsilon^2\lt\epsilon$, which is not true "for every $\epsilon\gt 0$". The argument is essentially fine, but if you want to be completely correct, you need to account for the cases where $\epsilon\gt 1$ awhere your asserted inequality fails, or explain why you can avoid considering them.

  • $\begingroup$ A sequence of sets converges if the lim inf and lim sup of the sequence are equal to some set A. $\endgroup$ – Damien Jun 23 '11 at 5:48
  • $\begingroup$ @Damien: In that case, Amitesh has provided you with an adequate comment on the proof of convergence; you've only established part of it. $\endgroup$ – Arturo Magidin Jun 23 '11 at 5:51

(1) Your proof of one implication is correct, namely, if $x$ is an element of either $\limsup A_n$ or $\liminf A_n$, then $x$ must be an element of $A_n$ for infinitely many $n$ and, in particular, $x$ must be an element of $A_n$ for some $n$. Therefore, $x\in A$.

However, the reverse inequality is "less trivial" in the sense that it uses the assumption $A_n\uparrow A$ and you should make sure you know why it is true. The reason is that if $x\in A$, then $x\in A_N$ for some positive integer $N$. In particular, $x\in A_n$ for all $n\geq N$ because of the assumption $A_n\uparrow A$. Therefore, $x\in \liminf A_n\subseteq \limsup A_n$.

(2) The analogous result is true. One direction, namely that $A\subseteq \liminf A_n\subseteq \limsup A_n$ should be clear. Conversely, if $x\in \limsup A_n$, you need to prove that $x\in A$. Do this by noting the fact that there is an increasing sequence of positive integers $n_1<n_2<\cdots$ such that $x\in A_{n_k}$ for all positive integers $k$. Conclude that $x\in A$.

(3) Your idea is correct. But you should understand (as Arturo points out) that your proof needs work if you would like to make it into a rigorous "epsilon-delta argument".


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.