# Tag Info

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In general, if $f:[0, \infty) \to [0, \infty)$ is a non-decreasing, concave function (i.e., $f(u + v) \leq f(u) + f(v)$ for all non-negative $u$ and $v$) vanishing only at $0$, then $d(x, y) = f\bigl(|x - y|\bigr)$ defines a metric on $\mathbf{R}$. Symmetry and positive-definiteness are obvious. The triangle inequality holds since for all real $x$, $y$, and ...

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Sure looks like it. It's translation invariant, so to prove the TE for $x \le y \le z$, adjust everything so that the lowest, $x$, of the three values is at $0$ (i.e., add $-x$ to all three numbers). Then you need to show that $$\sqrt{y} + \sqrt{z} \ge \sqrt{y+z}$$ for any nonnegative $y$ and $z$, which is true (by just squaring both sides).

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You need to deal with the fact that a rational centre has three rational co-ordinates - so, for example you can index the spheres not by ordered pairs, but by ordered quadruples $(x,y,z,r)$. You can adapt either method to deal with this. The first is possibly more straightforward.

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As $A\cap B \subset A$ , then any cover of $A\cap B$ by open sets is contained in a cover of A.

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I'm not quite sure to understand your statement $\forall r \in \mathbb Z, \exists s$ such that $q^{r+1} | p^s$ for negative $r$, so I'll assume it's only for non-negative integers (correct me if I'm wrong). Just looking at this statement and taking $r=0$ I see that if there exists $s_0$ such that $q^1=q | p^{s_0}$ then for $r \in \mathbb N$, $q^r | ... 3 Hint: a closed subset of a compact space is compact; a compact subset of a Hausdorff space is closed. 0 a) Openness is usually easier to prove than closedness. If a sequence$(a_n)$is not Cauchy, then there exists$\epsilon>0$such that for every$N$there are$n,m\ge N$with$|a_n-a_m|\ge \epsilon$. Now if$\|b-a\|_\infty<\epsilon/3$, then with the above indices$|b_n-b_m|\ge \epsilon/3$by the reverse triangle inequality. Thus, the set of all ... 2 Yes. Let$X$be a non-complete space. Consider a non-convergent Cauchy sequence$(a_n)$in$X$. Let$x\in X$be any point. The point$x$is not an accumulation point of$(a_n)$, because otherwise$(a_n)$(being Cauchy) would converge to$x$. Therefore there is an open neighbourhood$U$of$x$and an$N\in\mathbb N$such that$a_k\notin U$for$k>N$. Hence ... 1 It suffices to consider a Cauchy sequence that is not convergent and come to a contradiction. The set$S$consisting of the terms of the sequence is closed; otherwise some subsequence of it would be convergent, and for Cauchy sequences this would imply that the whole sequence is convergent. To make sure that$S$is not the whole space, delete from it one ... 1 Take all the continuous functions that satisfy$f(x +2\pi) = f(x)$for all real$x.$Make a distance on these by $$d(f,g) = \sqrt{ \frac{1}{\pi} \int_0^{2 \pi} (f(x) - g(x))^2 dx }$$ With this distance function, we can make an infinite set of functions that are distance$1$apart with $$f_n(x) = \frac{1}{\sqrt 2} \sin (nx).$$ The word "point" ... 1 This space can be defined using any set actually. It is generated by the metric called the discrete metric. Define$d: M \times M \rightarrow M$by $$d(x, y) = 0 \iff x = y \;\;\text{and }$$ $$d(x, y) = 1 \iff x \neq y$$ This is what is called the discrete metric and is defined for any set$M$. You can show for yourself that$d$actually constitutes a ... 1 Let$x\in X$and assume$f(x)\ne g(x)$. Since$Y$is Hausdorff, there are neighborhoods$U$and$V$of$f(x)$and$g(x)$respectively such that$U\cap V=\emptyset$. Since$f$and$g$are continuous,$f^{-1}(U)$and$g^{-1}(V)$are neighborhoods of$x$. Let$W=f^{-1}(U)\cap g^{-1}(V)$, which is a neighborhood of$x$. If$x'\in W$, then$f(x')\in U$and ... 0 fix$\epsilon=\frac{1}{1}$and find$y$of your definition of density. define$y_{1}:=y$fix$\epsilon=\frac{1}{2}$and find$y$of your definition of density. define$y_{2}:=y\cdots$fix$\epsilon=\frac{1}{k}$and find$y$of your definition of density. define$y_{k}:=y\cdots$this gives (one example for) your sequence. CONVERSE: by sequential ... 3 Suppose that$x^*\in X$is in the closure of$\{x\in X\,|\,f(x)=g(x)\}$. Then, there exists a sequence$(x_n)_{n\in\mathbb N}\subseteq X$converging to$x^*$such that$f(x_n)=g(x_n)$for all$n\in\mathbb N$. Using the triangle inequality for the metric on$Y$,$d_Y:Y\times Y\to[0,\infty)$, $$d_Y(f(x^*),g(x^*))\leq ... 3 The map X → Y × Y,\; x ↦ (f(x),g(x)) is continuous (why?) and, since the metric space Y is hausdorff, the diagonal Δ = \{(y,y);\; y ∈ Y\} is closed in Y × Y. 1 Let \{v_{i}\}_{i=1}^k be a set of k vectors in \mathbf{R}^n. By "sum of all pairs of inner-products", presumably you mean something like$$ \sum_{i<j} \langle v_{i}, v_{j}\rangle, $$and by "sum of Euclidean distances between all pairs" you mean$$ \sum_{i<j} \|v_{i} - v_{j}\|. $$Consider what happens with two vectors. Since you're asking about ... 0 So we have two definition. A Jordan curve is a subpace of the plane homeomorphic to the circle S^1. and A simple closed curve is a continuous map \gamma\colon [a,b]\to\mathbb{R}^2 such that \gamma(a)=\gamma(b) and for all c\in (a,b), if there exists a d\in[a,b] such that \gamma(c)=\gamma(d) then c=d. By the universal property of ... 0 To put it mathematically: A continuous map f\colon[a,b]\to\boldsymbol{R}^2 is called a Jordan curve iff f(a)=f(b) and f is injective on [a,b[. 0 Based on the flaws suggested in the comments this I think (IMHO) this is an easier way to approach some parts of the proof. To prove the line that x \in ∂X \implies x \in \overline A Suppose x is in the boundary of A and x is not in some closed set B which contains A. Then x \in B^c which is open and hence there is a neighbourhood V_xof ... 0 No, the first term \left | \left | x_{1}-x_{0} \right | - \left | y_{1}-y_{0} \right | \right | is not positive definite. In particular, d_{1}((x_{0},y_{0}),(x_{1},y_{1})) := \left | \left | x_{1}-x_{0} \right | - \left | y_{1}-y_{0} \right | \right | = 0 does not imply that (x_{0},y_{0}) = (x_{1},y_{1}). We can only get \left | x_{1}-x_{0} \right | ... 2 Another counterexample. Take any bounded metric space of diamter r. Any ball of radius greater than \frac{r}{2} will cover the entire space and so all such balls are equal no matter their center. 2 No. For example:$$X = \{ 1,2\}\\ d(x,y) = 0 \text{ if$x=y$, 1 otherwise}\\ x = 1, y= 2, r=16, s=26$$-2 Start with a nice discontinuous function f - something like the step function at 1/2. Find the Fourier series for this function. The series should converge to f in your metric (and not to any other continuous function, since the discontinuity is essential). But the sequence of partial sums is made up of continuous functions, and so is contained in ... 0 Let A\subseteq X be closed. For all n\in \mathbb N define$$U_n=\bigcup _{a\in A} B(a,\frac{1}{n}).$$U_n is open as a union of open balls. We prove that A=\bigcap _{n\in \mathbb N} U_n. Clearly A \subseteq \bigcap _{n\in \mathbb N} U_n. To prove A \supseteq \bigcap _{n\in \mathbb N} U_n we take x\notin A and show that x\notin \bigcap ... 1 Hints and clarification: (i) To show that T is continuous on (X,d) means to show that for any function f\in X=C[0,1] and any \epsilon>0, there exists \delta>0 such that if g\in X and d(f,g)<\delta, then d(T(f),T(g))<\epsilon. In particular, do NOT introduce a t_0. Here d is your supremum metric! (ii) Is the same, with a ... 2 As the OP corrected guessed the right subset to study is the set A\subset\ell^\infty consisting of all the sequences with zeros and ones, i.e., \{a_n\}\in A if and only if a_n\in\{0,1\}, for all n. Clearly if \{a_n\},\{b_n\}\in A and \{a_n\}\ne\{b_n\}, then$$ \|\{a_n\}-\{b_n\}\|_\infty=1, $$since for some n |a_n-b_n|=1. Also ... 0 Just apply the following lemma n times, starting with the empty set: Lemma. Given any closed set X\subseteq\mathbb R we can construct a closed set Y\subseteq\mathbb R such that X is the derived set of Y. In proving the lemma, it may be helpful to consider \mathbb R\setminus X as a union of disjoint open intervals. 2 Let X\subset \mathbb{R} be any set of reals containing only isolated points. Then we can define a function d:X\rightarrow\mathbb{R}_{>0} such that the intervals (x, x + d(x)) are mutually disjoint for all x\in X. The set$$ X'= \bigcup_{x\in X}\left\{x+d(x), x+\frac{1}{2}d(x), x+\frac{1}{3}d(x),\ldots\right\} $$has derived set X and contains ... 4 Draw ordinal \omega^{n-1} on real line. Here is picture of \omega^2. You can see that it's derived set is \omega and derived set of \omega is just a point. 0 I use the method... A \subset (X,d) and x_0 \epsilon X then by def U \subset X is called neighborhood of x_o if there exist G \epsilon \tau x_o \epsilon G \subset U as x_o is an accumulation point of A then G | {x_o} \bigcap A \neq \phi A \subset (X,d) and U \subset X then A \subset U Tell me if my method ... 0 It's easier to reason with balls in relation with open sets, since the definition of an open set is that it is one that contains an open ball around each of its points. So let us prove the contrapositive equivalence: p is in the complement of the closure of S iff there exists a ball centered at p which contains no point of S. I use the notation ... 3 \Rightarrow\quad by contradiction If p\in S the result is clear. Now assume p\not\in S and that all ball B centred at p doesn't contain any point of S then B^c is closed containing S and not p so their intersection doesn't contain p as well. Contradiction. \Leftarrow\quad by contraposition Assume p isn't in the closure of S then ... 0 If \varepsilon>0 and B=\left\{ x\mid d\left(p,x\right)<\varepsilon\right\} with S\cap B=\emptyset then B^{c} is a closed set with S\subset B^{c} so that \overline{S}\subset B^{c}. Consequently p\notin\overline{S}. If conversely p\notin\overline{S} then p\in\overline{S}^{c} wich is an open set. Then also \left\{ x\mid ... 0 To prove that P is in the closure of S, use the fact that any ball centered at p contains some point in S. Use this to build a sequence of points as you shrink the radius of the ball. Now you have a sequence of points in S who's limit is P. That should line up with your definition of a closed set. Something similar should get you the other direction. 4 Note that if (X,d) is metric space, then d'=d/(1+d) generates same topology of (X,d). So we only prove this proposition: Let (X_n,d_n) be a sequence of metric spaces, and d_n(x,y)\le 1 for all n and x,y\in X_n, then d((x_n),(y_n))=\sum_n 2^{-n} d_n(x_n,y_n) generates the product topology of X=\prod_n X_n. At first, we prove that for ... 0 This is already false for the graph which consists of a root and three nodes connected to it by edges. The proof follows from uniqueness of continuation of geodesics in Riemannian manifolds. 1 Assume that X is not connected. Then there exists a set F\subset X, such that F, K=X\smallsetminus F are non-empty, open and closed, and as they are closed they are compact. This means that$$ r=\mathrm{dist}(F,K)>0, $$and further there exist x\in F and y\in K, such that d(x,y)=r. In particular,$$\big(F\cup B(x,r)\big)\cap ... 1 I will use$B(x,r)$for the open ball centered at$x$with radius$r$,$\bar B(x,r)$for the corresponding closed ball and$\overline{B(x,r)}$for the closure of the open ball in$X$. First, notice that it is sufficient to prove that every closed ball is connected, because of $$B(x,r) = \bigcup_{s<r}\bar B(x,s).$$ (Remember that a union of a family of ... 2 Let$g = dx^2 + x^2dy^2$. First observe that for$(x,y)$,$(a,b) \in (\mathbb R^+ \times \mathbb R,g)$for all$n \geq 1:$$$d((x,y),(a,b)) \leq d((1/n,y),(x,y)) + d((1/n,y),(1/n,b)) + d((1/n,b),(a,b))$$ $$\leq x + 1/n|y - b| + a.$$ It follows that $$d((x,y),(a,b)) \leq x + a.$$ Now let$p_n = (x_n,y_n)$be a cauchy sequence in$\mathbb R^+ \times ...

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I started writing this as comments, but then ran out of space to give a satisfactory reply. (I rant too much to be confined to 400 characters!) At any rate, the answer is: no. But some clarification is necessary. I'm old and forgetful, so I will note John Baez describes the basic algorithm to geometric quantization fairly well. Examples Worth Considering ...

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Prove that $[\sum_{i=1}^{n} (x_i - y_i)^2]^{\frac12} \leq [\sum_{i=1}^{n} (x_i)^2]^{\frac12} + [\sum_{i=1}^{n} (y_i)^2]^{\frac12}$ holds for all $x_1, ..., x_n, y_1, ..., y_n \in \mathbb{R}$ and all positive integers $n$, then you can substitute $x_i$ with $x_i - z_i$ and $y_i$ with $y_i - z_i$ because they are real numbers too. Then take the limit as n goes ...

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Suppose the distance $d$ from $\pi$ to $\mathbb Q$ is positive. Then there is no rational number in the open interval with endpoints $\pi\pm d$, and thus no rational number smaller than the interval's length $2d$. Thus $2d>0$ is a lower bound of the set $\{1/n : n=1,2,3,\ldots\}$. Therefore $1/(2d)$ is an upper bound of $\mathbb N=\{1,2,3,\ldots\}$. ...

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Supposing this edit is correct, for the first question, for every $\varepsilon>0$ find a $q\in\mathbb Q$ so that $d(\pi,q)<\varepsilon$. For the second, find a point $a\in\mathbb R^+$ so that $d(-1,a)=1$ and prove that any point $b\in\mathbb R$ so that $d(a,b)<1$ must have $b\notin \mathbb R^+$.

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$B(0,1)$ usually means a unit ball centered at the point $\vec 0 = (0,0)$.

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B(0,1) means the unit ball with center in (0,0) and with radius 1. But the unit ball need not always be a circle! In Euclid (usual) metric space it is a circle, but there are other metrics in which it takes other forms (see unit balls for metric spaces generated by 1 and maximum norms).

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For $(a_n)$ the pointwise limit is $a(x)=x^2+3$ and we have $$||a_n-a||_1=\int_0^1 \left|\frac{-x^2}{n+1}+\frac{2}nx\right|dx\le\frac{1}{3(n+1)}+\frac{1}{n}\xrightarrow{n\to\infty}0$$ hence $$a_n\xrightarrow{||.||_1}a$$ by the same method prove that $$b_n\xrightarrow{||.||_2}(0,1,1)$$

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You need to decide whether $$\arctan|a+b| \leq \arctan|a| + \arctan|b|$$ Always holds. It is possible to show that this is indeed the case using (only) the fact that $\arctan(x)$ is a concave function on $[0,\infty)$.

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Pick a maximal subset $N$ of $X$ with the property that any two points in $N$ have distance at least $1$ from each other. Maximal means here that if $N\subsetneq N'\subseteq X$, then $N'$ contains points $n,n'$ with $n\ne n'$ and $d(n,n')<1$. That $N$ exists is an immediate consequence of Zorn's lemma. The point is that $N$ is a $(1,1)$-net, because if ...

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Any function $d$ is a metric if it satisfies the follow properties for all $x,y,z \in \Bbb R$: $d(x,y)=0$ $\iff x=y$ $d(x,y)=d(y,x)$ $d(x,y)\leq d(x,z) + d(z,y) \quad \forall x,y,z \in \Bbb R$

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Answer the following questions: Is it non-negative for all $(x,y)$? Is it zero if and only if $x = y$? Is it symmetric, i.e. $d(x,y) = d(y,x)$? Does it satisfy the triangle inequality, $d(x,z) \le d(x,y) + d(y,z)$? Questions 1 to 3 are quite easy to answer. Question 4 requires some basic algebra to answer.

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