Let $(A_n)$ be a decreasing sequence of sets. I am looking for an example that if $\mu^*(A_1) = \infty$, $$\mu^*(\cap_{n \in N} A_n)\not= \lim_{n\to \infty}\mu^*(A_n).$$  I saw from another  that if we define sets to be $A_1 = [0, \infty)$ and $A_n =\emptyset$ for $n \ge 2$, this sequence of sets can be an example which satisfies the above conditions. But, I do not understand why.   I think that $\lim_{n\to \infty}\mu^*(A_n)= 0$ as $A_n$ is empty set for $n \ge 2$. But, I also think that  $\cap_{n \in N} A_n = \emptyset$. That is, $\mu^*(\cap_{n \in N} A_n)=0$ as well.  Could you elaborate on this?
121971
I was wondering if someone could please give me an example of a sequence of decreasing sets where the first set has infinite Lebesgue measure; i.e., $\{B_{n}\}_{n=1}^{\infty}$ such that $m(B_{1}) = \infty$ but $m(\cap_{n=1}^{\infty} B_{n}) \neq \lim_{n \to \infty}m(B_{n})$?  thank you.
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.)
39556
I think I have the basic framework for this proof, but I am having trouble putting everything together in a convincing way.    I plan on showing that by the definition, an unbounded sequence $(a_n)$ has infinitely many $n$ such that $a_n&gt;M$ where $M&gt;0$, for some natural number $n$. I'm not sure how it is implied that there exists some $a_{n_k}$ such that $a_{n_k}&gt;M$ for some natural number $k$.  Let me know if I'm on the right track, and please help me string these ideas together!
106535
I am wondering if my proof is correct? Thank you for whoever willing to take a look at it for me.     Proof $\liminf E_k \subset \limsup E_k $   If $\{E_k\}_{k=1}^\infty$ is a sequence of sets, we define \begin{align*} \limsup E_k &amp; = \bigcap_{j=1}^\infty\left(\bigcup_{k=j}^\infty E_k\right)\\ &amp;= \bigcap_{j=1}^\infty(E_j \cup E_{j+1} \cup \cdots)\\ &amp;= (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots. \end{align*}  Therefore, $\limsup E_k$ consists of those points in $\mathbb{R}^n$ which belong to infinitely many $E_k$. \begin{align*} \liminf E_k &amp; = \bigcup_{j=1}^\infty\left(\bigcap_{k=j}^\infty E_k\right)\\ &amp;= \bigcup_{j=1}^\infty(E_j \cap E_{j+1} \cap \cdots)\\ &amp;= (E_1 \cap E_2 \cap \cdots) \cup (E_2 \cap E_3 \cap \cdots) \cup \cdots. \end{align*} Therefore, $\liminf E_k$ consists of those points in $\mathbb{R}^n$ which belong to all $E_k$ for $k \geq k_0$.  Now consider $x \in \liminf E_k$, then $x \in E_1 \cap E_2 \cap \cdots$, or $x \in E_2 \cap E_3 \cap \cdots$ and so on, that is to say, $x \in E_{k_0} \cup E_{k_0+1} \cup \dots$ for some $k_0$. More specifically, $x \in E_k \forall k \geq k_0$. Therefore, $x \in (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots$, so we showed $x \in \limsup E_k$.
112943
If $(a_n)$ is a decreasing sequence of strictly positive numbers and if $\sum{a_n}$ is convergent, show that $\lim{na_n}=0$  Since $(a_n)$ is decreasing and bounded below, by Monotonic Convergence Theorem, $(a_n)$ converges. So, there exists $N$ such that $|a_n-L|&lt; \epsilon$   Since $\sum{a_n}$ is convergent, by the Divergence test, we have $\lim_n{a_n}=0$, which means there exists $N$ such that $|a_n-0|&lt; \epsilon$  Then I get stuck at here.  I try to figure out the statement's meaning by inserting some examples, like $a_n=\dfrac{1}{n^2}$  Can anyone guide me on this question?
42409
Came across the following exercise in Bartle's Elements of Real Analysis. This is the solution I came up with. Would be grateful if someone could verify it for me and maybe suggest better/alternate solutions.  I also looked up these related questions - , ,  - but was not happy with proofs given there. I seem to need some help understanding these. Any such help is appreciated.      Show that every sequence in $\Bbb R$ either has a monotone increasing sub-sequence or a monotone decreasing sub-sequence.   Let $(x_n)$ be a sequence in $\Bbb R$. Suppose $(x_n)$ is not bounded. Without loss of generality we may assume that $(x_n)$ is not bounded above. Therefore given any real number there is a member of the sequence which is greater. Let $x_{n_1}$ be any member of the sequence.   There is $x_{n_2} \gt \sup\{x_1, x_2, ..., x_{n_1} \}$. For $i \gt 1$ let $x_{n_i} = \{x_1, x_2, ..., x_{n_{i - 1}}\}$ then $(x_{n_k})$ forms a monotone subsequence of $(x_n)$.   Now suppose instead that $(x_n)$ is bounded. By the Bolzano-Weierstrass Theorem there is a subsequence $(y_n)$ of $(x_n)$ which converges to a limit $y$. Without loss of generality there are infinitely many distinct values in $(y_n)$ that are unequal to $y$. Let $y_{k1}$ be the first such element. Let $y_{k2}$ be any element in $\{ y' \in (y_n) \ \ | \ \ |y' - y | \lt |y - y_{k1}|  \}$.   For $i \gt 1$ let $y_{ki} \in \{ y' \in (y_n)  \ \ | \ \ |y' - y | \lt |y - y_{k \ i - 1}|  \}$. Such $y_{ki}$ exists for every $i \in  \Bbb N$ since $ \lim (y_n) = y $. Now let $(y_{kn})$ be the sub-sequence of $(y_n)$ thus formed. At least one of the two following sets must contain infinitely many elements.    $\{ y \in (y_{kn}) \ \ | \ \ y \gt x\}$ $\{ y \in (y_{kn}) \ \ | \ \ y \lt x\}$   The one which does forms a monotone subsequence.
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 $L+\epsilon$.
97057
Proving that $$1_{\limsup A_n}=\limsup1_{A_n}$$  $$1_{\limsup A_n}= \begin{cases}1, x \in \bigcap_{n=1}^{+ \infty}\bigcup_{k=n}^{+\infty}A_n \\ 0, x \notin \ldots\end{cases}$$  or that $x$ is in infinitely many $A_n$. And also:  $$\limsup1_{A_n}=\inf_{n\geq1}\{\sup_{k \geq n}\{1_{A_n}\}\}$$ this function seems to be $1$ when $x\in A_n$ (just one set). I clearly am not understanding this $\inf\{\sup\{\ldots\}\}$ definition correctly. Can someone enlighten this for me. I have a test very soon.
39075
Show that if $A_1\subseteq A_2\subseteq A_3\subseteq\cdots$ is an increasing sequence of measurable sets (so $A_j\subseteq A_{j+1}$ for every positive integer $j$), then we have $$m\left(\bigcup_{j=1}^\infty A_j\right)=\lim_{j\to\infty}m(A_j)$$  Here is my proof: According to the $\sigma$-algebra property, $\bigcup_{j=1}^{\infty}A_j$ is a measurable set, so it makes sense to talk about $m(\bigcup_{j=1}^{\infty}A_j)$. Firstly, I prove that $\lim_{j\to\infty}m(A_j)\leq m(\bigcup_{j=1}^{\infty}A_j)$. This is because for any given positive integer $N$, $A_N\subseteq \bigcup_{j=1}^{\infty}A_i$, according to monotonicity, we have $m(A_N)\leq m(\bigcup_{j=1}^{\infty}A_i)$. Take the limit,we will have $\lim_{j\to\infty}m(A_j)\leq m(\bigcup_{j=1}^{\infty}A_j)$. Secondly, I prove that $m(\bigcup_{j=1}^{\infty}A_j)\leq \lim_{j\to\infty}m(A_j)$. For any given positive integer $N$, $\bigcup_{j=1}^N A_j = A_N$. According to monotonicity,we have $m\left(\bigcup_{j=1}^N A_j\right)=m(A_N)\leq \lim_{j\to\infty}m(A_j)$. Take the limit, we will have $m\left(\bigcup_{j=1}^\infty A_j\right) \leq \lim_{j\to\infty} m(A_j)$. Combine the above two arguments, we will see that $$m\left(\bigcup_{j=1}^\infty A_j\right)=\lim_{j\to\infty} m(A_j)$$$\Box$  The above is my proof, unlike many books, my proof does not use the property of countable additivity. So I doubt my proof is correct. Who can point out where are my mistakes?
104250
Let's consider the  defined by:      Take $ A\subset \mathbb N$; define the sequence $x_n= \dfrac{|A\cap[1,n]|}{n}$, and then if $\lim\limits_{n\to\infty} x_n$ exists, call it $d(A)$, the density of $A$ over $\mathbb N$.   My teacher said that the collection of sets that have a natural density is not an algebra, but I don't know why. I can't even find a set that does not have a natural density.  Please help me )=
62760
what is the number of subsets of the set {k∈N|1≤k≤n} with no two consecutive numbers? The answer says: $$a_n=a_{n-1}+a_{n-2}$$ with the starting conditions: $$a_0=1, a_1=2$$1. why does $a_1=2$? $$$$ 2. How does this recursion was created, I understand that if $a_n$ mean all the places with no consecutive numbers so, $a_{n-1}$ come from the fact that after the first place all those places satisfy this condition, but why there is need to "look" at what happen after the two places?
8283
Let $E$ be an empty set. Then, $\sup(E) = -\infty$ and $\inf(E)=+\infty$. I thought it is only meaningful to talk about $\inf(E)$ and $\sup(E)$ if $E$ is non-empty and bounded?   Thank you.
60526
Prove that if a sequence satisfies $\lim\limits_{n\to\infty}|a_{n+1} - a_n|=0$ then the set of its limit points is connected.   My professor once mentioned a proposition likewise but I cannot find the exact version of it. Can anyone help?
93622
I am trying to prove some theorem, and in the proof of it I have to check the fact that if the sequence $\{a_n\}_{n\in \mathbb{N}}$ converges, then $$\lim_{n\to \infty} n(a_n - a_{n-1}) = 0. $$ I think that this is true because I have tried with many different sequences and it holds, but I do not know how to prove it.  If this is not true, what is a counterexample for it?
43686
Does there exist a decreasing sequence  of sets ($A_{n+1}\subseteq A_n$) ; $\{A_n\}$ of real numbers such that   $m^*(A_n)&lt;\infty , \forall n \in \mathbb N$ and $m^*(\cap_{n\in \mathbb N} A_n) &lt; \lim_{n\to \infty} m^*(A_n)$ ? Here $m^*$ is the Lebesgue outer measure on real line
76069
Let $(X,\tau)$ be a topological space, $\emptyset\neq A\subseteq X$ and $(x_n)$ be a sequence in $A$. I could easily show that any limit of $(x_n)$ belongs to $\overline{A}$. My note says that if $x\in\overline{A},$ $x$ is not necessarily a limit of a sequence in $A$. But I cannot find a counterexample. Now in my attempt to find a counterexample, which I failed, I could conjecture that if the topological space is metrizable then, if $x\in\overline{A}$ then $x$ is a limit of a sequence in $A$. Is my conjecture true? And how would you find a counterexample to the original problem? Please help. Thanks.
16166
I'm having trouble to prove the following question:  Supose $\{A_n\}_{n\in\mathbb{N}}$ is a family of sets such that $A_1\subset A_2\subset A_3\subset\ldots$ (it's possible to have $A_n=A_{n+1}$). I need to prove that $$\lim_{n\to\infty}A_n = A_1\cup\bigcup_{n=2}^\infty(A_{n}\backslash A_{n-1}).$$  I'm not sure if this is useful, but I proved that $A_{n}\backslash A_{n-1}$ and $A_{n'}\backslash A_{n'-1}$ are disjoint if $n\neq n'$.  Thank you.