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Let me start by saying you are right in a sense, and I think the article is at best being unclear, but the larger point that the equation of ellipse cuts out a sphere is also right. There are a few things going on, hence a long answer below. What one is trying to describe is the topology of a set defined by a quadratic equation $\frac{x^2}{a^2}+\frac{y^2}{... 1 Your$\Sigma$does not satisfy condition (a): if you take the subset$\emptyset\subset\Sigma$, then the union$\bigcup\emptyset=\emptyset$is not an element of$\Sigma$. That is, the union of elements of$\Sigma$consisting of no elements at all is not in$\Sigma$. 4 For the Baire category proof. Use the OP proof up to $$C^{'}=\displaystyle\bigcup_{i=1}^{\infty} \partial B_{i} .$$ Then note$C^{'}$is a complete metric space, written as a countable union of sets$\partial B_i$, which are closed (in$C'$) sets with empty interior (in$C'$). 3 To show no such cover via open balls exists is easy: any such cover$\mathcal{O}$must involve at least 2 open balls (since the radii are finite). So pick$O\in\mathcal{O}$, and let$A=O$,$B=\bigcup (\mathcal{O}\setminus\{O\})$. Then$A$is open by assumption, while$B$is open as a union of open sets; and$A\sqcup B=\mathbb{R}^n$. But this contradicts ... 0 The following answer is not rigorous. To make it rigorous you may need to know more about smooth manifold.$\mathbb S^3 = \{ (u, v) \in \mathbb C^2 : |u|^2 + |v|^2 =1\}$is itself a$3$-manifold, which means that locally it looks like an open sets in$\mathbb R^3$. Mathematically speaking, it means that for each point$p\in \mathbb S^3$, there is a ... 1 There is a map from$T^3$to$SO(3)$, called "Euler angles" -- basically, you use each angular coordinate to specify "yaw", "pitch", and "roll". This map is, however, singular, much like the map from$T^2$to$S^2$discussed in the comments. The singularities are often referred to by the generic name "gimbal lock". You can read about this in books on ... 2 Here's another one, in$\mathbb R$.$X = $the set of midpoints of the complementary intervals of the Cantor set. So every point of$X$is the midpoint of an interval with no other points of$X$; that is every point is isolated. But$X'$is the Cantor set, uncountable. 2 I'm not sure what you're doing with your approach. Let me sketch another one: For each$x\in X,$set$r_x = d(x,X\setminus \{x\}).$Because each$x\in X$is an isolated point of$X,$each$r_x>0.$Show that the collection$\{B(x,r_x/2): x \in X\}$is pairwise disjoint. Note that each$B(x,r_x/2)$contains an element in$\mathbb Q^n.$5 Unfortunately, what you’re trying to prove is false. Let $$X=\left\{\left\langle\frac{2m+1}{2^n},\frac1{2^n}\right\rangle:n\in\Bbb N\text{ and }m\in\Bbb Z\right\}\;;$$ every point of$X$is isolated in$X$, but$X'$is the whole$x$-axis. Added: For that matter, you could just as well use the simpler set $$\left\{\left\langle\frac{m}{2^n},\frac1{2^n}\... 0 Hint: all point of X are isolated i.e there exists open balls B_x(x,r_x>0) with B(x,r_x)\cap B(y,r_y) is empty if x\neq y. Take an element q_x\in Q^n\cap B(x,r_x) an define f:X\rightarrow Q^n f(x)=q_x. Use the fact that Q^n is numerable and f is injective. Remark. A point x is isolated if there is a ball B(x,r_x) such that B(x,r_x)\... 2 As you try this on points farther and farther along the way, you start approaching the point * from the other side. To be continuous also there, you'd have to drag * itself around the circle once, and then the points slightly past it once and a bit. But that's in contradiction to your rule that you only move them until you hit * the first time. So your ... 1 it is not continuous at *. You construct an homotopy H_t:S^1\rightarrow S^1 such that H_t(x)=c_t(x) where c_t is the path going towards * from x clockwise. You must have H_t(*)=*. There exists a sequence lim_nx_n=* anticlockwise such that x_n\neq * and lim_nH_t(x_n)=c_t(x_n)=* anticlockwise otherwise H_t(S^1), t<1 is homeomorphic to ... 4 It seems to me that in the usual polar coordinates on the circle, OP is talking about the homotopy$$ H: S^1 \times [0, 1] \to S^1 : (\theta, s) \mapsto (1-s) \theta. $$For s > 0, this map is not continuous at \theta = 0, because no neighborhood of 0 maps to a small neighborhood of H(0, s). Hence it's not a deformation retraction from S^1 to ... 2 You have a family (\gamma_t)_{t\in[0,1]} of continuous maps from the circle to itself such that \gamma_0 is the identity, \gamma_1 is constant, and \gamma_t\to\gamma_1 pointwise is t\to1. But contractibility requires that \gamma_t\to\gamma_1 uniformly. At least that's the way I read your description. It's sufficiently vague that it could be the ... 0 Let U=\prod_{m=1}^\infty (a_m,b_m)\cap \mathbb R^\infty be an arbitrary set of a canonical base of the box topology. Since for each n the intersection U\cap\mathbb R’^n is \prod_{m=1}^n (a_m,b_m) \times\prod_{m=n+1}^\infty \{0\} iff a_m<0<b_m for each m>n and is empty, otherwise, we see that the set U is open in \mathbb R^\infty. ... 1 I think you're working far too hard. Pulling X=\mathbb{R}^2 apart works well! Define f\colon X\to\mathbb{R}^2\setminus\{ [-1,1]\times(-\infty,\infty)\} by f(x,y)=(x,y) if |x|>1, f(x,y)=(x+4,y) if |x|\le 1. Let's call Y=\mathbb{R}^2\setminus\{ [-1,1]\times(-\infty,\infty)\}. This isn't open because the image of the infinite vertical strip ... 0 As the user Nobody pointed out, the map F is continuous as it is a composition of continuous maps: The map \iota: [0,1] \rightarrow [0,1], t \mapsto 1-t, and the Homotopy G. Then the map F = G \circ (Id_X , \iota), is continuous as both G and \iota are contiuous. 1 r(t) : t \mapsto 1-t is a continuous function [0,1] \to [0,1] (1_X, r) is thus a continuous function X \times [0,1] \to X \times [0,1] F \circ (1_X, r) is thus a continuous function X \times [0,1] \to X 1 Note that each one of the three balls is relatively open AND CLOSED in K, being the intersection of a closed ball of \mathbb{C} with K. For instance, B_1 = \overline{B_1} \cap K, where the closure is meant in \mathbb{C}. So you have found a subset of K (namely, B_1) which is relatively closed and open in K, hence K is disconnected. Your ... 0 Trigonometric functions are circular functions. Complex cosine is: cos z = (e^(iz) + e^(-iz))/2 where z = x + iy where x and y are real numbers and z is a complex number. Suppose we restrict z so modulus z = 1 = modulus (x + iy). Hence, the image of z is the unit circle in the complex plane. What if we were to look at the map X = (x, y)? Let X ... 2 The function f(z) = (z+1/2)^2 will do the job. It maps the upper half of the circle bijectively to a curve that starts in the first quadrant, moves left to the fourth quadrant, comes down and intersects the negative real axis, continues downward and rightward under 0 and then comes back up to end at 1/4. The bottom half of the circle gets matched to ... 1 Take any two points \exp(i\alpha) \ne \exp(i\beta) on the unit circle, and a nonconstant analytic function f on a neighbourhood of the circle such that f(\exp(i\alpha)) = f(\exp(i\beta)). Then the curve \gamma(t) = f(\exp(it)), 0 \le t \le 2\pi intersects itself (\gamma(\alpha) = \gamma(\beta)). If you want the curve to be regular, require the ... 0 I think$$cos(x)+i sin(2x)$$is a good, simple example. (x=\frac{lnz}{i}) 1 The essential property is that given a compact metric space X and r>0 there is a finite number of balls covering X. Let 0<\lambda<1 and make the simplifying assumption that X may be covered by two closed balls B_0 and B_1 of size 1. Then assume (again for simplicity) that each of \Lambda_0=B_0 and \Lambda_1=B_1 (both are compact ... 3 Suppose towards contradiction that \mathbb{R}=U\sqcup V, U and V nonempty and open. Then there are real numbers x\in U and y\in V - without loss of generality let's assume x<y. Now - looking at the interval [x, y] - let$$z=\inf\{a\in [x, y]: a\in V\}.$$Such a real z exists, by the completeness of \mathbb{R}. Now, which piece of \... 1 The general outline is to write your given compact metric space A as a decomposition into two sets, then four sets, then eight sets, etc., in the exact same fashion that the Cantor set is decomposed, except that whereas the decomposition elements of the Cantor set are pairwise disjoint, the decomposition elements in A need not be pairwise disjoint. Be ... 1 Let X be a compact metric space. The key point is that we can subdivide X into smaller and smaller regions, such that at each stage there are only finitely many regions total. Specifically, we want for each n a partition$$X=Y^n_1\sqcup . . . \sqcup Y^n_{k_n}$$of X such that each Y^n_i has size (that is, diameter) less than (say) 2^{-n}. If we ... 2 Unfortunately, you are not free to morph the two halves in the very restricted fashion described in your question. You simply do not have that much control over the three given sets. Those three sets can each be much wilder and off kilter than you are imagining. If your proof were correct, then the given plane P that you found, passing through the centers ... 1 A misinterpretation I think you have misunderstood the stated fact in the linked answer. When it is stated that given any basis of a topological space, you can always find a subset of that basis which itself is a basis, and of minimum possible size. what is meant is the following given any basis of a topological space, you can always find a subset ... 1 T_0 is not enough. For n\in\Bbb N let U_n=\{k\in\Bbb N:k\ge n\}, and let \tau=\{\varnothing\}\cup\{U_n:n\in\Bbb N\}; \tau is a T_0 topology on \Bbb N. Let$$f:\Bbb N\to\Bbb N:n\mapsto n+1$$and$$g:\Bbb N\to\Bbb N:n\mapsto\begin{cases} 0,&\text{if }n=0\\ 1,&\text{if }n\ge 1\;. \end{cases}$$Check that \langle\Bbb N,\tau\rangle is ... 0 Suppose this is false and take x \in S, where S is the subset of non isolated points. Let y_0 \in S and take a neighborhood U_1 of x which does not contain y_0. Now take y_1 \in S, y_1 \in U_1 (this is clearly possible because every neighborhood of x contains infinite points of S), and U_2 \subset U_1 such that y_1 \notin U_2. Take ... 0 A very general result is true; I’ll leave just part of the proof to you. We’ll need the following result, which is proved here. Let \langle X,\le\rangle be any linear order, and let \tau be the order topology on X. Then the space \langle X,\tau\rangle is T_5 (i.e., completely normal and T_1). Now let \langle X,\le\rangle be a linear order,... 1 Your conclusion is false. At least part of it. For any convergent sequence \{a_n\} with limit q in \mathbb{R}^n, if we let B=\{a_n; n\in \mathbb{N}\}, then \overline{B}=B\bigcup\{q\}, regardless of whether or not it is on the boundary of an open set (every point in \mathbb{R}^n is on the boundary of some open set). However, your second ... 1 Given the function f:\overline{B}_r(p)\to\overline{B}_r(p)\setminus\{p\}, let$$g:\overline{B}_r(0)\to\overline{B}_r(0)\setminus\{0\}:x\mapsto f(x+p)-p\;,$$and apply the original result to g. 1 The following should answer your question unless I made a mistake (it's sketchy in a couple places), and is probably a bit overkill but it clarifies some relevant concepts: Let X be path-connected, let a,b\in X, and let \Omega_{[a,b]}(X) denote the space (with the compact-open topology) of continuous paths \gamma\colon [0,1]\to X such that \gamma(0)... -1 Here's another example that uses the idea of infinite oscillation as suggested by @YCor: On \mathbb R, let f(t) be a C^\infty function that is 0 for t<0, that strictly increases to 1 on [0,1], and that is 1 for t>1. The oscillatory piece: Define$$g(t) = \begin{cases}\sin(1/(t(1-t))\exp (-1/(t(1-t)), & 0< t < 1 \\ 0, &... 3 I assume the sequences are of abelian groups. The first one splits since the right term is a free abelian group. The second one does not necessarily split, since one can take$A=\mathbb Q$and$B$the quotient. Since the middle term is torsion free, the sequence cannot split. 1 I assume the groups involved here are commutative. The first sequence splits because$Z^2$is a free abelian group. if$f:B\rightarrow Z^2$,$f(u)=e_1,f(v)=e_2,$where$e_1,e_2$are generators of$Z^2$, write$g(e_1)=u, g(e_2)=v$. 0 See Henno Brandsma’s answer for a way to express your idea much more clearly. A slightly different approach is to let$\mathscr{B}$be a countable base for$X$, and let $$\mathscr{B}_0=\{B\in\mathscr{B}:|B\cap A|\text{ is countable}\}\;,$$ the family of basic open sets that contain at most countably many points of$A$.$\mathscr{B}_0$is a subset of the ... 2 Along the same lines: For the first question,$E_i$is nowhere dense iff$\overline{E_i}$is nowhere dense (both saying that$\overline{E_i}$has empty interior). And if$X$is not a countable union of$ \overline{E_i}$then it is not a countable union of$E_i$either. For the second question:${\mathbb R} = (\mathbb R \setminus {\mathbb Q} ) \cup {\...

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First, a comment: I think you are implicitly assuming that every set is either open or closed. This is not true: consider e.g. the rationals! On to your questions: For the first one, there is no restriction on the type of sets involved: they may be open, closed, or neither, so long as they are nowhere dense. (Although in fact it's not hard to prove the ...

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You essentially seem to be having the right idea, but the proof is very unclear. Start by picking a countable base $\{B_n: n \in \mathbb{N}\}$ of $X$. For every $x \in A\setminus A'$ ($A' =$ the set of limit points of $A$, that you call $D$) we can pick a basic open set $B_{n(x)}$ such that $B_{n(x)} \cap A = \{x\}$. If $x, y$ in $A \setminus A'$, then if ...

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More or less ok, but you should perhaps pay more attention as to the choice of the sets $B_a$? Let $B_n$, $n\geq 1$ be a countable basis. For $a\in A\setminus D$ there is an element $B_{n(a)}$ which does not intersect $A\setminus\{a\}$ and since $a\in B_{n(a)}$ the sets $B_{n(a)}$, $a\in A\setminus D$ must be disjoint. So the map $a\in A\setminus D \mapsto n(... 2 Your last paragraph is basically already the answer to your question. Suppose$A$is hollow. Then for any (nonempty) open set$O$,$O$must contain some$x\in \mathbb{R}\setminus A$(why? otherwise$O$would be a subset of the interior of$A$, which would make$A$not hollow). But this is exactly the statement that the complement of$A$is dense! So ... 3 Yes - assuming the axiom of choice, cardinalities are well-ordered. In fact, this statement is equivalent to the axiom of choice. So your argument does indeed work - assuming the axiom of choice. To polish it up a bit, here's how I'd write it: Let$S$be the set of cardinalities of bases of$X$. By AC, the cardinalities are well-ordered, so$S$has a ... 1 That$A_i \cap A_j \not= \emptyset$is not always true. Consider for each$n \in \mathbb{Z}$:$A_n:=\{n\}\times \Bbb{R}$as a subspace of$\Bbb{R}^2$and let$A=\mathbb{R}\times \{0\}$. Then$A_n\cap A \not= \emptyset$for each$n$, but$A_n \cap A_m = \emptyset$whenever$m\not=n$. Let$X$be a topological space. Then, show that every continuous function$...

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Hint that should make it clear how to proceed: The connected component that contains $A$, in $A\cup\big(\bigcup_{i\in I} A_i\big)$, contains a point in each $A_i$ by your non-empty intersection condition. If some connected component contains a point in $A_i$, then can it happen that it does not contain all of $A_i$?

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Urysohn's Lemma implies that in a normal space you can separate points and closed sets by continuous functions; in fact you can separate disjoint closed sets. But what you're trying to do is impossible in general. Not every Hausdorff space is normal, and what you're trying to do is not always possible in normal spaces. If $C$ and $f$ are as in your post ...

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Let $f$ be a $C^\infty$ function on $\mathbf{R}$ that is 2-antiperiodic ($f(x)+2=-f(x),$ $\forall x$) and such that $f(x)=\sin(1/x)\exp(-1/x)$ for all $x\in ]0,1]$. Let $g$ be a $C^\infty$ function on $\mathbf{R}$ that is 2-periodic and such that $g(x)=x$ for all $x\in]0,1]$. Note that $x\mapsto \exp(i\pi x/2)$ is 4-periodic and injective on $[0,4[$. ...

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It seems that I have got it: It is necessary to establish that $U = U_q \setminus {U_p}^-$ is an open set and thus a neighborhood of $x_0$. It is not said explicitly in the proof.

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