Given the rectangle $Q$, find a subset $S$ of $Q$ whose closure equals $Q$, such that $S$ contains at most one point on each vertical line and at most one point on each horizontal line.  I really don't figure out why there must exists such a set; can you give me an example?
40473
Given a subset $A$ of $\mathbb{R}^{n}$, a point $x \in \mathbb{R}^{n}$ is said to be in the boundary of A if and only if for every open rectangle $B\subseteq\mathbb{R}^{n}$ with $x\in B$, $B$ contains both a point of $A$ and a point of $\mathbb{R}^{n}\setminus A$.  My question is from Spivak's Calculus on Manifolds: Construct a set $A \subseteq [0,1]\times [0,1]$ such that $A$ contains at most one point on each horizontal and vertical line but has boundary equal to $[0,1]\times[0,1]$.
112303
I am trying to show that closure of a compact set in a Regular space is compact, but I am hitting some hurdles as a compact set in a regular space need not be closed.  This is what I started off with: If $ C $ is any compact subset, let $ U $ be any open neighbourhood containing $ C $. Then by regularity of $ X $, $ \forall x \in C, \exists V_x : \overset-V_x \subseteq U $ where $ V_x $ is a neighbourhood of $ x $. So $ \lbrace  V_x:x \in C\rbrace $ is an open cover for C, which by compactness of C , simplifies to $ \lbrace  V_{x_i}:x_i \in C\,\,for\,i=1,...,n\rbrace $. Now I end up looking at $ \cup \overset-V_{x_i} $ which  should contain $ \overset-C $. But I get stuck here. Should I look at the intersection rather??
117041
Let (X,d) be a compact metric space. For every open cover, show there exists ε > 0 such that for every x ∈ X, B(x,ε) is contained in some member of the cover.   My attempt:  (X,d) is compact. Therefore there exists a finite subcover of X.  Any element x in X must lie in some member of the cover, say x ∈ Ui. Otherwise they would not constitute a cover.  Since Ui is open, by definition every point is interior, so there exists ε > 0 such that B(x,ε) is contained in Ui.   I haven't used the fact the subcover is finite, or the fact X is a metric space rather than just topological space, so I feel my reasoning is flawed.   Any help is greatly appreciated!
43169
How many ways can a rectangle be partitioned by either vertical or horizontal lines into n sub-rectangles? At first I thought it would be:        f(n) = 4f(n-1) - 2f(n-2)   where f(0) = 1     and f(1) = 1    but the recurrence relation only counts the cases in which at least one side (either top, bottom, left or right of the original rectangle) is not split into sub-rectangles. There are many other partitions that don't belong to those simple cases like  [EDIT ImageShack has removed the picture. One of the cases is the sixth partition when n = 4 in the picture in the accepted answer below.]  Any other related problem suggestions are welcome. Also it is nice to know how to traverse this partition efficiently.
71765
I would be very glad if you helped me solving this problem. It's about topology. Okay let's begin :    Let $A$ be a non-empty open subset of $\mathbb R$. Let $x_0 \in A$ and $B = [x_0,\infty) ∩ A^c$ (complement of $A$)  First I should prove that $B$ is a non-empty, closed set. Then I should prove that :     $B$ is bounded below.   $\inf B &gt; x_0$   $\inf B$ doesn't belong to $A$.   $[x_0, \inf B) \subseteq A$.    Then I should explain how to deﬁne a number $\beta$ so that $(\beta,\inf B) \subseteq  A$, where $\beta$ doesn't belong to $A$.    I really can't get up with proofs for these question. Can you please help me solving this or at least give me some hint or sketch? I would be very grateful. Thank you :)
96677
There is a set A which contains some discrete points (1-dimension), for example {1, 3, 37, 59}. And I want to pick one point from A which can minimize the sum of distances between this point and others.  For this "1-D line", many people say the median is the right point, but could anybody give me a concrete proof on it?
116873
I'm looking for an algorithm to detect if two rectangles intersect (one at an arbitrary angle, the other with only vertical/horizontal lines).  Testing if a corner of one is in the other ALMOST works.  It fails if the rectangles form a cross-like shape.  It seems like a good idea to avoid using slopes of the lines, which would require special cases for vertical lines.
53345
Proposition : The only sets in $R^n$ which are both open and closed are the empty set and $R^n$ itself.  I came up with the proof of this claim and I'd like to know if my proof is correct.  Proof:  Let $S$ be the nonempty proper subset of $R^n$ which is both open and closed. Then $T = R^n-S$ is also a nonempty proper subset of $R^n$ which is both open and closed. Now take a point $s_0$ in $S$ and $t_0$ in $T$. Then we can consider a closed box $J_1$ which contains both $s_0$ and $t_0$.  The n-dimensional interval $I_1^k$ is defined by the inequalities   $s_k\le x_k \le t_k$ (k=1, 2, $...$, n ), where each $s_k$ and $t_k$ are the k-th coordinates of $s_0$ and $t_0$. Without loss of generality, assume that for all k, $s_k \le t_k$.  Here $J_1$ denotes the catesian product   $J_1$ = $I_1^1$ x  $\cdots$ x $I_n^1$   Now, bisect all of the n 1-dimensional intervals and the point $\frac {s_0+t_0}{2}$ will lie in the center of the bisected box. This point belongs to either $S$ or $T$. Say it belongs to $S$ then, rename it $s_1$ and the original $t_0$ to be $t_1$. Now again consider a box that has both points as vertices like the previous case and repeat the process.   Writing $I_k^m$ = [$s_k^m$,$t_k^m$],   we have  $t_k^m$ - $s_k^m$ = $\frac {t_k-s_k}{2^{m-2}}$  Then for each fixed k, the sup of all left endpoints must therefore be equal to the inf of all right endpoints, and their common value we donote by $b_k$ (actually we know this by Bolzano-Weierstrass).   Now we assert that b=($b_1$,$...$,$b_n$) is a limit point of $S$ and $T$. To see this, take any n-ball B(b; $r$). The point b of course belongs to each of the boxes $J_1$, $J_2$, $...$ constructed above, and when $m$ is such that $max_k\frac {t_k-s_k}{2^{m-2}}$ &lt; r/2, this neighborhood will include $J_m$ and since $J_m$ contains both points of $S$ and $T$, b is a limit point of both sets. But since both sets are closed, they both contain b, which is a contradiction. Hence, the proposition is true.
66011
I've been experimenting with recursive sequences lately and I've come up with this problem:     Let  $a_n= \cos(a_{n-1})$ with $a_0 \in \Bbb{R}$ and $L=[a_1,a_2,...,a_n,...].$   Does there exist an $a_0$ such that $L$ is dense in $[-1,1]?$     I know of $3$ ways of examining whether a set is dense:  $i)$The definition, that is, whether its closure is the set on which it is dense, in our case this means if: $\bar L=[-1,1]$.  ii)$(\forall x \in [-1,1])(\forall \epsilon&gt;0)(\exists b \in L):|x-b|&lt;\epsilon$ $iii)$ $(\forall x \in [-1,1])(\exists b_n \subseteq L):b_n\rightarrow  x$   So far I haven't been able to use these to answer the question. I tried plugging in different values of $a_0$ and see where that leads but I have not found any corresponding promising "pattern" for $a_n$. Any ideas on how to approach this?
76077
There is a point $(x,y)$, and a rectangle $a(x_1,y_1),b(x_2,y_2),c(x_3,y_3),d(x_4,y_4)$, how can one check if the point inside the rectangle?
11200
I need to find the closure of these sets on the ordered square: $$A = \{(1/n)\times 0 | n\in \mathbb{Z}_+\}$$ $$B = \{(1-1/n)\times \frac{1}{2}| n\in \mathbb{Z}_+\}$$ $$C = \{x\times 0 | 0&lt;x&lt;1\}$$ $$D = \{x\times \frac{1}{2}| 0&lt;x&lt;1\}$$ $$E = \{\frac{1}{2}\times y | 0&lt;y&lt;1\}$$ I know that the closure is just the set itself united with its limit points. For $A$, the set of limit points, I imagine, are $\{&lt;1,0&gt;\}$, that is, the $x$ coordinate is $1$ and the $y$ is $0$. But  is says the reverse. What am I getting wrong? Following the same reasoning, $B' = \{&lt;1,\frac{1}{2}&gt;\}$ $C' = \{&lt;0,0&gt;,&lt;0,1&gt;\}$ What am I doing wrong?
76995
This is exercise $1.5$A from Engelking's book, page $47$.  Verify that $X$ is $T_{0}$ if and only if $\overline{\{x\}} \neq \overline{\{y\}}$ for every pair of distinct points $x,y$.  My try:  Suppose first that $X$ is $T_{0}$ and let $x,y$ be two distinct points of $X$. Then since $X$ is $T_{0}$ wlog we can find an open set $U$ such that $x \in U$ and $y \not \in U$. Let us show that $\overline{\{x\}} \neq \overline{\{y\}}$. Suppose otherwise, then $x \in \overline{\{x\}} \subseteq \overline{\{y\}}$. This in turn implies that $U \cap \{y\} \neq \emptyset$ so that $y \in U$, a contradiction.  For the other direction, let $x,y$ be distinct points of $X$. Then either $\overline{\{x\}}$ is not contained in $\overline{\{y\}}$ or $\overline{\{y\}}$ is not contained in $\overline{\{x\}}$. Assume the former, then $x \not \in \overline{\{y\}}$ so we can find an open set $U$ such that $U$ contains $x$ and $U$ does not intersects $\{y\}$. If the other case happens we can find an open set $V$ such that $V$ contains $y$ and $x \not \in V$. Therefore $X$ is $T_{0}$.   Is this OK?
61114
I have a set with $N$ members $\{1,2, \dots, N\}$. I would like to know number of set partitions in which each subset is either of size $1$ or $2$, i.e., cardinality of each subset in the partition is maximum $2$. For example with $N=3$, we have the set partitions of maximum size $2$ are $\{1,2,3\}, \{(1,2),3)\}, \{1,(2,3)\}, \{(1,3),2\}$. However I am not interested in $\{(1,2,3)\}$ since it is of length $3$. Kindly tell me number of such set partitions for a set with $N$ members.
85905
I want to prove $\Leftarrow$.  I just need one more step, now. If I find a compact set that contains K, then it will finish the proof. Rudin take a k-cell to do that, but before it he proves that every k-cell is compact, and I think I can take a more simple compact to do this step. I has thinked in the closed ball, but don't know how to prove that it is compact. Can anyone help me to find this set, if it really exists? Thanks.
46729
Let $X$ be a Hausdorff space and $A\subset X$.  Define $A'=\{x\in X\mid x\text{ is a limit point of }A\}$.  Prove that $A'$ is closed in $X$.  Relevant information:  (1.) Every neighborhood of a point $x\in A'$ contains a point $y\in A'$ distinct from $x$ (in fact, in a Hausdorff space, every neighborhood of $x$ contains $\infty$-many points of $A$ distinct from $x$)  (2.) In a Hausdorff space, every sequence has a unique limit.  On first glance, it should be an easy proof, but I've made little progress.  I was planning on showing $\overline{A'}=A'$.  Firstly, $A'\subset \overline{A'}$ trivially.  To show $\overline{A'}\subset A'$, proceed by contradiction.  Assume there exists $x\in\overline{A'}$ such that $x\not\in A'$.  This should yield an easy contradiction but I don't see it.  In particular, I'm unsure if the fact that $x\in\overline{A'}$ implies that there actually exists a sequence in $A'$ converging to $x$.  By (2) we know all sequences have unique limits, but do we know that elements in the closure are limits of sequences?  If this is true, it should yield an easy contradiction.  Any help?
36933
Let $I$ be a special rectangle in $\mathbb{R}^n$, and denote $\lambda(A)$ the measure of $A$. Prove that the following conditions are equivalent:  a) $\lambda(I)=0$  b) $I^{\circ}=\emptyset$ (i.e., the interior of $I$ is empty)  c) $I$ is contained in an affine subspace of $\mathbb{R}^n$ having dimension smaller than $n$. (An affine subspace is any set of the form $\{x_0+x|x\in E\}$, where $x_0\in\mathbb{R}^n$ is fixed and $E$ is a subspace of the vector space $\mathbb{R}^n$. Its dimension is equal to the dimension of $E$.)  The implication $a\implies b$ isn't too bad (using the definition if $I=[a_1,b_1]\times...\times[a_n,b_n]$ then $\lambda(I)=(b_1-a_2)...(b_n-a_n)$. So we conclude $a_i=b_i$ for some $i$. And since $I^{\circ}$ is open, there can't be anything contained in it). I also see how c) makes sense (at least intuitively), but not sure how to show it formally.  Thank you.
61084
What is the number of rectangles that can be obtained by joining four of the 12 vertices of a 12-sided regular polygon?  The thing that I've tried is to find out at what intervals should I connect to vertices of the polygon so that a rectangle can be formed. But that didn't seem to help at all .