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Suppose that $(y_\beta)_{\beta \in B}$ is a subnet of $(x_\alpha)$ via the map $h:B \rightarrow A$, and this subnet converges to $x$. Let $O$ be an open subset of $X$ that contains $x$. Then by the convergence we have some $\beta_0 \in B$, such that for all $\beta \ge \beta_0$ we have that $(y_\beta) \in O$. Then consider the set $A_O=\{ \alpha \in A: ... 0 If you require the limit to be unique (as you said, like if we assume$Y$to be Hausdorff), then$g$must be continuous. First, the function$g$is defined everywhere, since the limit of$f$exists and is unique everywhere. Now you have to show that the limit of$g$exitst (and is unique, but that's implied by the Hausdorff assumption) and that it matches ... 0 If$X$is a standard flat torus and the equivalence relation declares two points to be equivalent if they lie on the same line with a (fixed) irrational slope, the quotient is clearly not metrizable. 1 If$(X,d)$is a pseudo-metric space, and if you define: $$x \sim y \iff d(x,y) = 0$$ then$(X/_\sim, d^*)$is a metric space, with: $$d^*([x],[y]) = d(x,y)$$ It has to be checked that$d^*$is well-defined. The space$(X/_\sim, d^*)$is called the metric identification of$(X,d)$. If I remember well, this is an exercise in Stephen Willard's General Topology. ... 1 You may show If$A$and$B$is a separation of$X−Y$, then$Y∪A$and$Y∪B$are connected. by assuming$A$and$B$is a separation of$X−Y$and$Y∪A$OR$Y∪B$is not connected and then derive a contradiction. You do not show above statement by assuming$A$and$B$is a separation of$X−Y$and$Y∪A$AND$Y∪B$is not connected and deriving ... 1 Consider the set of all open sets containing just one particular point in Euclidean space. If you like, call that point the origin, so we're talking about the set of all open sets that contain the origin. Their intersection contains just that one point. That's not an open set. An arbitrary union of open sets includes as a subset some open neighborhood of ... 2 Why is that.. only finitely many intersections of open sets is open? In the context of metric spaces, this is because: Every finite set$X$of positive real numbers has a positive lower bound. An infinite set$X$of positive integers may not have a positive lower bound. For example,$\{\frac{1}{1+n} : n \in \mathbb{N}\}$is a set of positive numbers ... 3 Perhaps this would be easier if it is motivated from the metric space context. If$(X,d)$is a metric space, a subset$U \subset X$is open if for all$x \in X$, there is some ball$B(x,r)$of some radius$r > 0$, around$x$such that$B(x,r) \subset U$. Now suppose we have a family of open sets$U_i$for$i \in I$some indexing set. The matter is ... 7 Generally this goes back to the underlying concept of convergence of a sequence and the so called Hausdorff'sche Umgebungsaxiome (see: Felix Hausdorff. Grundzüge der Mengenlehre. Veit & Comp., Leipzig, 1914). Intuitively the convergence of a sequence to a point means, that in each neighbourhood of the point can be found almost all parts of the ... 0 Another way is to make use of the characterisation of compactness by families of closed sets with the finite intersection property: A topological space$X$is compact iff whenever$\mathcal{A}$is a family of closed subsets of$X$with the finite intersection property (i.e., the intersection of any nonempty finite subfamily of$\mathcal{A}$is nonempty), ... 0 The space you are concerned with is called the Arens–Fort space. For these questions, the important facts about this space are the following: Any subset of$E$which does not contain$(0,0)$is open. Any neighbourhood of$(0,0)$is actually clopen (= closed and open). If$N$is any neighbourhood of$(0,0)$and$m > 0$, then$N \setminus ( \{ m \} \times ...

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I assume that you assume a neighbourhood to be open, otherwise i am not sure if your definition determines a topology. $E$ is not locally compact: First observe that for each $(m,n) \neq (0,0)$ the set $\{(m,n)\}$ is open, since by definition of your topology $\{(m,n)\}$ is a neighbourhood of $(m,n)$. Now consider an arbitrary neighbourhood $U$ of ...

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Let's start with along the lines of the standard proof. Let us divide $[0,1]$ into $2k$ intervals of length $1/k$; i.e. $[0,1/k]$, $[1/k,2/k]$, $[2/k,3/k]$, etc. Now by Dirichlet principle there are two numbers $a\ne b$ such that $\{a\alpha\}$, $\{b\alpha\}$ which are in the same interval. If $b>a$, then $(b-a)$ is a positive integer and either ...

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This is alright, essentially you use that $SO(2)$ is homeomorphoic to $S^1$. A similar method is possible for $SO(n)$, but the parametrization is a bit more involved. An inductive proof by continuously transforming one column to the standard base vector (using precisely this $n=2$ case) works best, I suggest.

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Compactness is key. For the real line, keep in mind the Heine-Borel theorem which says a set is compact iff it is closed and bounded. As a counterexample then, take an unbounded closed set $\mathbb{R}$ (in the space $\mathbb{R}$), and take $\mathbb{Z}$.

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For $q \in \mathbb Q$, let $\mathcal O_q = (\mathbb R \setminus \mathbb Q) \cup \{q\}$ i.e. the set of all irrational numbers and $q$. Then, $\mathcal O_q$ is cocountable and $\mathbb R = \bigcup_{q\in \mathbb Q} \mathcal O_q$ but no finite subcover would cover all of the rational numbers.

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In a Hausdorff space, compact sets are closed. To see this, suppose $K$ is compact, and let $x\notin K$. Then for each $k\in K$, consider an open ball $U_k$ around $k$ and an open ball $V_k$ around $x$ which are disjoint. The collection of all such $U_k$ is a cover of $K$. Take a finite subcover, and the intersection of the corresponding $V_k$ is an open set ...

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Assume $X$ is Hausdorff. If every subspace is compact then every subspace is closed. So the topology is discrete. Now take the cover by singletons. If this has a finite subcover then the space must be finite.

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Using Hausdorff to show $B$ is closed: $Y$ is a Hausdorff space then the set $D=\{(y,y)\in Y\times Y\ ,\ y\in Y\}$ is closed in $Y\times Y$. Now let $h:\overline{A}\to Y\times Y$ the map defined by $h(x)=(g_1(x),g_2(x))$ it is a continuous map, and $B=h^{-1}(D)$ is closed.

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if $A$ is already closed then there is nothing to prove. so we may suppose that there is a point $a$ in the boundary of $A$ such that $g_1(a) = b_1$ and $g_2(a)=b_2$ where $b_1$ and $b_2$ are distinct points. since $Y$ is Hausdorff we may choose disjoint open neighbourhoods $V_1$ and $V_2$ of $b_1$ and $b_2$. and we must have: $$a \in U = g_1^{-1}(V_1) ... 1 Take the space \mathbb{R}. Then the subset \mathbb{R}_{>0} of positive elements is a subset of \mathbb{R}, but is not a union of connected components of \mathbb{R} (the only connected component of \mathbb{R} is \mathbb{R} itself). But then that's okay, because \mathbb{R}_{>0} isn't clopen in \mathbb{R} (it is only open). What you seem ... 2 How can we find such an example? Since f is a bijection, we may assume wlog that X and Y are the same set, just with different topologies (and then f is just the identity map). The fact that f is continuous then means that every Y-open set is also an X-open set. To prevent f from being a homeomorphism, the converse should not hold, i.e. not ... 1 Hint for (c) \implies (d): apply Zorn's Lemma to a given collection of open subsets ordered under inclusion. This gives that (a) \implies (b) as well, since you've already shown (a) \iff (c) and (b) \iff (d). A comment on your proof of (b) \implies (a): take care with the definitions of maximality and minimality. You write Suppose (b) holds ... 1 wspin's counterexample: Identify the sphere with the inscribed cube and let A be the front, bottom, and back faces. Let B be the top, left, and right faces. Then each connected component of A\cap A^\star is disjoint from its image under the antipodal map, and similarly for B \cap B^\star. Assume X \subset A. Then X\subset A\cap A^\star, so that ... 1 For a more direct approach: Suppose that I_1 , I_2 , \ldots are open intervals in \mathbb{R} such that I_n has length 3^{-n}. Note that I_1 must be disjoint from either [ 0 , \frac 1 3 ] or [ \frac 2 3 , 1 ] (or both, I guess, if I_1 was chosen particularly badly). Let us label one of these intervals disjoint from I_1 by A_1. Now I_2 ... 2 Hint: For (a), suppose that F= G_1\cup G_2 for G_1 and G_2 proper closed sets. If G_1 and G_2 are irreducible, we are done. If not, WLOG assume G_1 is reducible. Write G_1=H_1 \cup H_2 for H_1 and H_2 proper closed subsets. Continue in this way, expressing a reducible closed set from the previous step as a union of proper closed sets, if a ... 4 Lemma. If X \subseteq [0,1] is a strong measure zero set, and f : [0,1] \to [0,1] is continuous, then f[X] is also strong measure zero. proof. Note that f is uniformly continuous, so for each \varepsilon > 0 there is a \delta ( \varepsilon ) > 0 such that if I \subseteq [0,1] is an interval of length \leq \delta ( \varepsilon ), ... 0 For instance the Sierpinski space (S, \tau) with S = \{0,1\} and \tau = \{\emptyset, \{0\}, S\} is quite useful, see http://en.wikipedia.org/wiki/Sierpi%C5%84ski_space . Also, with this kind of "minimal sandbox" topology you can learn about closures, continuous maps, and so on without having to deal with infinite sets. 1 Note that if U \subseteq \mathbb{R} is nonempty and open in topology \tau_B, then \operatorname{Int}_{\text{met}} ( U ) (the interior of U with respect to the usual metric/order topology on \mathbb{R}) must also be nonempty. (You can see this by just looking at the sets in the base: if a < b, then \operatorname{Int}_{\text{met}} ( [a , b) ) = ... 0 The space is separable because the set of all rational numbers is a countable dense set. Every basic open set, and therefore every nonempty open set, contains a rational number. 4 In general, when we write f: A \to B, we mean f \subseteq A \times B, with the condition that if (x,a), (x,b) \in f, then a = b. Since the second entry in the ordered pair is uniquely determined by x, we call this element f(x). This condition is just saying that f is well-defined. We often think of f as some machine that takes an element ... 2 The ordered pair is not f it is the graph of f. So f: \mathbb{R} \to \mathbb{R} and G: \mathbb{R} \to \mathbb{R}^2 where G is the graphing function. 4 I would say f:\mathbb{R}\rightarrow \mathbb{R}. You could consider the related function g:\mathbb{R}\rightarrow\mathbb{R}^2 that is defined as \{(x,(x,f(x))\mid x\in\mathbb{R}\} 2 First, your "counterexample" doesn't work. Take the \epsilon-ball with \epsilon=1 about your single point space, X=\{p\}. Then B_1(p) = \{ x \in X \;|\; d(x,p)<1 \} = \{p\} (since d(p,p)=0<1). Therefore, \{ p \} is open. Just as in the subspace topology. So why do the subspace topology and the "usual" topology match? It essentially boils ... 1 I believe they're talking about the 'natural' topology of a graph as a realisation of a complex whose 0-cells are the vertices of the graph, and whose 1-cells are the edges of the graph. 2 On objects it is simply the underlying function p:E \to B. On arrows, i.e. on paths, it is given by composition: given [f]:x \to Y homotopy class of a path in \Pi(E), you define \Pi(p)([f]):=[p \circ f]. Finally, this functor is a covering of groupoid thanks to the well-known fact that a covering admits a lift for any path in the base space, after ... 1 Your main issue is to get around the positivity of n. So lets change the standard proof. By the pigeon hole principle you can find some m \in \mathbb Z so that mx = k+ y with k \in \mathbb Z and y \in (0 , \frac{1}{k}). If m >0, you are happy. If m <0, then show that there exists some l \in \mathbb N so that ly \in (\frac{k-1}{k}, ... 0 Let S = \{\{n \alpha\} | n \in \mathbb{N} \}. First note that it is enough to show that S intersects (0,1/N) for arbitrarily large N \in \mathbb{N}. By the usual pigeonhole argument, for some k \in \mathbb{N}, \{k\alpha\} is in either (0,1/2N) or (1-1/2N,1). Now, in the first case we are done. In the second case, let \{k\alpha\} = ... 1 Pick any k\in\mathbb{N}. By the pigeonhole principle, there are two multiples of \alpha whose fractional part lie within 1/k of each other. Taking the difference, there is a multiple of \alpha with (positive) fractional part <1/k. It follows that every x\in [0,1] is within 1/k of some \{n\alpha\}, for any k. 1 Injectivity loos rather correct. For the second part, you essentially want to "invert" polar coordinates restricted to the unit circle. The map \phi^{-1} provides the angle \varphi such that x=\cos \varphi and y = \sin \varphi. Now,$$ \varphi = \phi^{-1}(x,y) = \operatorname{atan2}(y,x),  where atan2 is defined as in this Wikipedia page. The ...

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A fat Cantor set. A is closed, its interior is empty, so $A$ minus its iterior is $A$ itself. See previous question: http://math.stackexchange.com/a/287872/442

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What about $(\Bbb R-\Bbb Q)\cap [0,1]$

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Something that you could employ in answering this is to note that every regular CW-complex is in fact triangulable. It turns out that every closed topological manifold of dimension other than four is homeomorphic to a CW-complex [Kirby-Siebenmann, On the triangulation of manifolds and the Hauptvermutung]. On the other hand Ciprian Manolescu has recently ...

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Your proof is accurate and well written. I see no need for improvement.

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I want to mainly mention a point that hasn't been made in the other answers so far. If $X$ is any topological space and $x \in X$ has a finite neighborhood basis, then it has neighborhood basis of size one. This is because the intersection of any finite family of neighborhoods of $x$ is also a neighborhood of $x$, and is a subset of each of those ...

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Each of your finitely many $U_i$ is a neighbourhood of $x$, hence by definiiton contains an open ball $B_{r_i}(x)$ for some positive number $r_i$. Pick one for each $i$. Let $r$ be the minimum of these finitely many positive real numbers, so $r$ is itself a positive real number. Consider $B_{r/2}(x)$. As $B_r(x)\setminus B_{r/2}(x)$ is nonempty, we have thus ...

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a hint: no. consider n=2 and the matrices over $R$. The determinant of the matrices is 1 or -1. These are the two conected componets (matrices with determinant 1 and matrices with determinant -1) .

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but can we dispense with continuity and surjectivity? Yes. Your - correct - proof uses only the closedness of $p$ to show that every $y\in Y$ has an open neighbourhood $W_y$ such that $p^{-1}(W_y)$ is covered by finitely many of the $U_\alpha$, and since $Y$ is covered by finitely many $W_y$, it then follows that $X$ is covered by finitely many ...

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What does it mean for $(x,y)$ to be in the boundary of $A\times B$ ? Can you show that either for every open $x\in U\subseteq X$ and $y\in V\subseteq Y$ there are $a\in A\cap U$ and a $b\in B\cap V$, and there is a $u\in U\setminus A$ or for every open $x\in U\subseteq X$ and $y\in V\subseteq Y$ there are $a\in A\cap U$ and a $b\in B\cap V$, and ...

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Here is what I mean: If $C$ is clopen subset of a space $X$, then $C=\cup_{x\in C}C_x$, where $C_x$ is the connected component in $X$ containing $x$. It is clear that $C\subset \cup_{x\in C}C_x$. Let $x\in C$, and we want to show that $C_x\subset C$. Now we take $A=C_x\cap C$ and $B=C_x\cap C^c$, we have $A\cup B=C_x$ and $A\cap B=\emptyset$ and $A$, $B$ ...

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