Show that if a non-empty bounded set A $\subset$ R and s = sup A then there is a sequence such that A converges to s. I guess that I should use the epsilon definition of supreme for this question but I am not exactly sure how.
123494
Let $S$ be a bounded set. Prove there is an increasing sequence ($s_n$) of points in $S$ such that $\lim s_n = \sup S$. Note: If $\sup S$ is in $S$, it’s sufficient to define $s_n = \sup S$ for all $n$.  What I have tried so far:  Let $M=\sup S, L = \lim\, s_n,\,|s_n-L|&lt;\epsilon/2$, then $M-\epsilon &lt; L- \epsilon/2 &lt; s_n &lt; M &lt; M+\epsilon$. But I get this based on $M &lt;= \epsilon+L$, because $\epsilon+L$ is an upper bound for $s_n$.   However, as $s_n$ are just points in $S$, so I don't know if it's sufficient to say $M$, which is the $\sup$ of $S$, is smaller than $L+\epsilon$.
110843
I know that there are at least 3 other questions with the similar problem. I hope, though, you won't flag this as a duplicate since I've got some specific questions here that haven't been answered there before.  So we all know the theorem states:  The sequence of functions $\{f_n\}_{n\in \mathbb{N}}$ converges uniformly to the function $f$ if and only if $$\lim_{n\to \infty}\sup{|f_n(x)-f(x)|} = 0$$  And the proof that I've found says the following:  For the right hand side implication:  Let $l_n = \sup{|f_n(x)-f(x)|}$. From the assumtion (that the function converges uniformly) we know that:  $$(\forall\epsilon&gt;0)(\exists n_0 \in \mathbb{N})(\forall x \in X)(n \geq n_0 \Rightarrow |f_n(x)-f(x)|) &lt; \frac{\epsilon}{2}$$  Therefore $l_n \leq \frac{\epsilon}{2} &lt; \epsilon \Rightarrow \lim_{n \to \infty}l_n = 0$. Proof done.  Here's what confuses me:  The fact that $l_n$ is the supremum of the set with elements $|f_n(x)-f(x)|$ tells us that it is always greater than those elements.  How does that tell us that it's less than $\epsilon$. The only justification that I can come up with is to recall that the supremum is the least upper bound, and since $\epsilon$ is also an upper bound in this case, than the supremum must be less than it. But still not sure how valid that is.  Another thing: If my justification is correct, why do we even use $\frac{\epsilon}{2}$? Can't we just immediately conclude that $l_n$ must be less than $\epsilon$. It seems unnecessary.   And finally, the textbook I'm reading this from is missing the proof for the left hand side implication. So could anyone give me some hints on how to prove that side?  Thanks.    EDIT:  Some similar questions I've found:    There was one more, I think, but I can't find it anymore.
117279
Let $1&lt;p&lt;\infty$, $(u_n)_{n\in\mathbb{N}}\subseteq l^p(\mathbb{N})$ such that $u_n$ converges weakly to $u\in l^p(\mathbb{N})$. Prove that   (1) $(u_n)$ is bounded and (2) for all fixed component $j\in n\in\mathbb{N}$ it is $\lim_{n\to\infty} u_{n,j}=u_j$ (in lecture we called it "pointwise convergence of sequences").   My questions are: (1) is "$(u_n)$ bounded", i.e. $\|u_n\|&lt;\infty$ for all $n\in\mathbb{N}$ already satisfied because $(u_n)_{n\in\mathbb{N}}\subseteq l^p(\mathbb{N})$?   For (2) it is to show that $|u_{n,j}-u_j|\to 0,\; n\to\infty$ for fixed $j\in\mathbb{N}$. But I don't know how to do this.
112407
Let $A,B\subseteq\mathbb{R}$ two non-empty sets such that $a&lt;b$ for all $a\in A$ and $b\in B$.   Which of the following properties hold:  (1) $\sup A&lt;\infty$  (2) $\sup B&lt;\infty$  (3) $\sup A&lt;\inf B$  (4) $\sup A=\inf B$ iff there exist sequences $(a_n)_{\mathbb{N}}$ in $A$ and $(b_n)_{\mathbb{N}}$ in $B$ such that $\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}y_n$  (5) $\sup A&lt;\inf B$ iff there exist sequences $(a_n)_{\mathbb{N}}$ in $A$ and $(b_n)_{\mathbb{N}}$ in $B$ such that $\lim\limits_{n\to\infty}x_n&lt;\lim\limits_{n\to\infty}y_n$  My ideas:  (2) is false:  Consider $B=\{x+y: x,y\in \mathbb{R}_{&gt;0}\}$ and $A=\{0\}.$ Then $A$ and $B$ satisfy $a&lt;b$ for all $a\in A$ and $b\in B$, but $\sup B$ does not exist in $\mathbb{R}$.  (3) My attempt: Fix $b\in B$, by assumption it is $a&lt;b$ for all $a\in A$. Thus, $b$ is an upper bound for $A$, therefore $\sup A&lt;b$. Or follows $\sup A\le b$ (I'm not sure at this point)? Since $\sup A$ is a lower bound for $B$, $\sup A&lt;\inf B$...or is $\sup A=\inf B$ as well possible?  (1) is correct: $\sup A\le \inf B$ and $\inf B$ exists in $\mathbb{R}$. (Suppose $\inf B$ doesn't exist in $\mathbb{R}$. $\inf B=-\infty$ only can happen if $B$ is not bounded below which is a contradiction to $B$ is bounded below by a non-empty set $A$.)   (4) Suppose $\sup A=\sup B$. Take $(a_n)_{\mathbb{N}}$ in $A$ converging to $\sup A$ and $(b_n)_{\mathbb{N}}$ in $B$ converging to $\sup B$, for the existence see here  . Then: $\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}y_n$. However, I don't know about the other direction.  (5) I don't know..  Is everything what I have tried correct? How does it work for (3) and the remaining directions in (4) and (5)? Thank you
5433
The problem is the following: suppose $s_1&gt;s_2&gt;0$ and let $s_{n+1}=\frac{1}{2} (s_n+s_{n-1})$ for $n\geq 2$. Show that ($s_n$) converges.  Now, here is what I figured out:   $s_2&lt;s_4$: Base Case for induction that $s_{2n}$ is an increasing sequence. Assume $s_{2n-2}&lt;s_{2n}$. Induction step: $s_{2n}&lt;s_{2n+2}$ $s_1&gt;s_3$: Base Case for induction that $s_{2n-1}$ is a decreasing sequence. Assume $s_{2n-1}&lt;s_{2n-3}$. Induction step: $s_{2n+1}&lt;s_{2n-1}$.   I have proved those two. However arguing in favor of convergence has me going around in circles. Since $s_1&gt;s_2$ and (as discovered during the formulation of Base Cases) $s_3&gt;s_4$, I figured it might be a good idea ot assume that if every odd member of the original sequence ($s_n$) is greater than the following even member, then the limit would be somewhere in between, the two (odd and even) sequences won't cross. Hence the upper and lower bounds would be $s_1$ and $s_2$ respectively. Here is how I approach this:   Assume $s_{2n-1}&gt;s_{2n}$ Show that $s_{2n+1}&gt;s_{2n+2}$   The proof as I mentioned has me running in circles. Any assistance?
2274
I'm stuck with the proof of the following:  Suppose that $(s_n)$ converges to $s$, $(t_n)$ converges to $t$, and $s_n \leq t_n \: \forall \: n$. Prove that $s \leq t$.   I've tried starting with $s_n \leq t_n \: \forall : n$ and the definitions of each limit (i.e. $|s_n - s| \leq \epsilon \: \forall \: n &gt; N_1$), but I'm not really getting very far. Any help is appreciated!
44461
I would appreciate your help!  How can we show that if a sequence of positive real numbers $a_n$ converges to $a\in\mathbb{R}$ with $a&gt;0$, then $(a_0 a_1\cdots a_n)^{\frac{1}{n}}$ converges to $a$.
6578
If $A,B$ non empty, upper bounded sets and $A+B=\{a+b\mid a\in A, b\in B\}$, how can I prove that $\sup(A+B)=\sup A+\sup B$?
112089
Let $(a_n)$ be a sequence such that $(a_n+a_{n+1})$ is convergent. Prove that $(a_n/n)$ converges to $0$.   $(a_n+a_{n+1})$ convergent to $L$ means that for all $\epsilon$, there exists $N$ such that for all $n\geq N$, $|L-a_n-a_{n+1}|&lt;\epsilon$, or in other words $L-\epsilon&lt;a_n+a_{n+1}&lt;L+\epsilon$. How to proceed from here?
76540
Given a sequence of sets, is there some well-defined notion of a limit of a set?   In other words, given some universe set $U$, I am wondering if there is a topology on $2^U$ (the powerset of $U$) such that the usual intersection and the union limits converge in that topology.  As an explicit example, let $U=\mathbb{N}$,  $S_n = \{x\in \mathbb{N} | n&lt; x \le 2n \}$,   $T_n = \{n\}$.  The limit of both sequences above should be the empty set by the following argument:  \begin{align} S_n &amp;\subset (n,\infty) \\\\ \lim_{n\to\infty} S_n &amp;\subset \lim_{n\to\infty} (n,\infty) = \cap_{n\in\mathbb{N}} (n,\infty) = \emptyset  \end{align}  (I'm not sure how to justify passing a set inclusion to the limit.)
28137
I would like to prove the following statement.  Let A and B be nonempty disjoint subsets of $\mathbb{R}$ where A is closed and B is compact. Then there exists $a \in A$ and $b \in B$ such that $inf${$|a-b|$:$a \in A, b \in B$}=$|a-b|$    I read a similar question above where the concept of "limit point" and others were used, but my class (Introduction to Analysis) has not gone that far; most of all, I would like to prove the statement from the knowledge that the class covered.  So far, I have learned definition of limit of a sequence, Bolzano-Weierstrass Theorem. Also, I just learned (and proved) that there exists a sequence ${a_n}$ in E such that $\lim{a_n} = \inf{E}$ if E is a nonempty subset of $\mathbb{R}$ (whether E is bounded below or not due to the extended real numbers).   From the things that I learned, first, I constructed a set $E$={$|a-b|$:$a\in{A}, b\in{B}$} and I know that there exists a sequence ${x_{n}}$ in E such that $\lim{x_{n}}=infE$. Since ${x_{n}} \in{E}, x_{n}=a_{n}-b_{n}$ where $a_{n} \in A, b_{n} \in B$. Let $infE = \beta$. Then, this leads to $\lim{(a_{n}-b_{n})}=\beta$. Then, for any $\epsilon$>0, there exists $N \in$ $\mathbb{N}$ such that $|a_{n}-b_{n}-\beta|&lt;\epsilon$ for all n $\ge$ N.   Then, I could not proceed. I should use conditions that A and B are disjoint, A is closed, and B is compact. Could anyone help me proceed? I had a feeling that using contradiction might work from the link above, but it is quite tough. Since I am a novice (just started taking Introduction to Analysis), I would very much appreciate it if you could help me prove this at the beginner level.
82309
Let $A$ and $B$ be nonempty sets of positive real numbers that are bounded above. Also let $AB = \{ab: a \in A, b \in B \}$. Prove that $AB$ is bounded above and $\sup(AB) = (\sup A) (\sup B)$.   So $\sup A$ and $\sup B$ exist by completeness. An upper bound for $AB$ is $(\sup A)(\sup B)$. Let $\alpha = \sup A$ and $\beta = \sup B$. We want to show that if $c$ is an upper bound for $AB$ then $\alpha \beta \leq c$. For $a \in A$, $ab \leq c$ for all $b \in B$. So $c/b$ is an upper bound for $A$. Thus $\alpha \leq c/b$. It follows that $\alpha \beta \leq c$.  Is this correct?
62673
I have a question that says "Show that there is a bounded sequence $x_n$ which is not convergent but has the property that $x_n - x_{n+1} \to 0$ as $n \to 0$.  What does this mean? Do I need to come up with an example or does the problem actually want me to prove such proposition?  By the way, I see that this sequence looks like Cauchy because of $x_n - x_{n+1} \to 0$ as $n \to 0$, but it is obviously not.
56679
a) Prove that if $\sum a_n$ converges absolutely and $b_n$ is a bounded sequence, then also $\sum a_nb_n$ converges absolutely.  I wanted to use the comparison test to show it's true, but I think I got something mixed up. Here's what I have so far.  $b_n$ is bounded $\Rightarrow \exists c &gt; 0: |b_k| &lt; c, \forall k \in \mathbb N$  $\sum a_nb_n &lt; \sum a_nc &lt; c\sum a_n$. I'm very tempted to use the comparison test here and say, as $\sum a_n$ converges absolutetly, then so does $c\sum a_n$, but for that I needed the inverted relation, right? I would need $\sum a_n &gt; c\sum a_n$, which is not true. However isn't it obvious that something absolutely convergent multiplied by a constant is also absolutetly convergent? Is there a mathematical way to write this? Thanks a lot in advance.  b) Refute with a counter example: if $\sum a_n$ converges and $b_n$ is a bounded sequence, then also $\sum a_nb_n$ converges.  Is this even possible? I mean, it has to be, but it doesn't make sense to me. If something converges, it means it's bounded, right? And I thought, well, since bounded + bounded = bounded, then bounded*bounded would also get me something bounded again.  Anyway, my idea would be to use for $a_n$ an alternating series that ist only convergent, for example $(-1)^n \frac 1 {n^2}$. But something tells me I'm trying to prove $a_n b_n$ is not absolutely convergent.  Thanks a lot in advance guys!
64428
The question is based on the following exercise in real analysis:     Assume that $A,B\subset{\Bbb R}$ are both bounded and $x&gt;0$ for all $x\in A\cup B$. Show that $$ \sup(AB)=\sup A\sup B $$   where    $$ AB:=\{ab\in{\Bbb R}:a\in A, b\in B\}. $$   Since $0&lt;a\leq\sup A$ and $0&lt;b\leq\sup B$ for all $a\in A$ and $b\in B$, we have $$ ab\leq\sup A\sup B $$ for all $ab\in AB$ which implies that $\sup AB\leq\sup A\sup B$. I have trouble with another direction: $$ \sup AB\geq\sup A\sup B $$ I was trying to show that for every $\epsilon &gt;0$, $\sup AB-\epsilon \geq \sup A\sup B$. If one uses the definition of supremum, one has the estimates that for every $\epsilon&gt;0$, $$ \sup A-\epsilon\leq a, \quad \sup B-\epsilon\leq b $$ for some $a\in A,\ b\in B$. It follows that $$ \sup A\sup B\leq (a+\epsilon)(b+\epsilon)=ab+\epsilon(a+b)+\epsilon^2\leq \sup AB+\epsilon (a+b)+\epsilon^2 $$ which seems quite close to what I want. How can I go on?
107521
I am assuming my teacher wants us to conclude they converge by being bounded and monotone, since part (i) is to show that $a_n$ is monotone increasing and $b_n$ is monotone decreasing.  I am confused though because I have never worked with a sequence defined with... another sequence.
37443
I am taking a course in Real Analysis this semester and thought I would work ahead a little bit.  I am just reading for the moment, but came across an interesting exercise, and was wondering if I could see how one of you would solve it.  Assume that $A$ and $B$ are nonempty, bounded above, and satisfy $B⊆A$. Show $\sup B≤\sup A.$  Not so much that the question perplexes me, but rather I wanted to see an example of a rigorous proof for this question.