# Tag Info

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What you have written is not true. You need to show that $J(x)$ is non-empty for every $x\in X$. But suppose if $x_0\in X$ is such that $\|x_0\| < 1$ and if possible let there exist $f\in J(x_0)$. Then $\|f\| = \|x\| < 1$. But $\|f\| = \sup_\limits{x\in X, x\neq 0}\frac{\|f(x)\|}{\|x\|} \geq \frac{\|f(x_0)\|}{\|x_0\|}= \frac{\|x_0\|}{\|x_0\|} = 1$. ...

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Your answer to part (a) is correct. As Rolf Hoyer noted in the comments, you can use the point $x'$ from part (a) to answer part (b): you’ll find that no matter how small a $\delta>0$ you choose, $B_{\tilde\rho}(x',\delta)\setminus U(x,\epsilon)\ne\varnothing$, because, for instance, $x_k'+\frac{\delta}2>x_k+\epsilon$ for all sufficiently large $k$. ...

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Hint: So, this is how I picture your sets: and I added my $B_y$. For any $y$: $$B_y=\overline{\bigcup_{n=1}^{\infty}\{(\frac{1}{n},y)\}}={\{(0,y)\}\cup\bigcup_{n=1}^{\infty}\{(\frac{1}{n},y)\}}.$$ Then $$\bigcup_{y\in[0,1]}B_y\cup A_0 \cup \{(0,0)\}=A \cup [(0,0), (0,1)].$$ If, however, the following is the right interpretation: then ...

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Let $X=\Bbb R\setminus\Bbb Q$. Fix an $\alpha\in X$, e.g., $\sqrt2$, and let $D=\alpha\Bbb Q=\{\alpha q:q\in\Bbb Q\}$. A set $U\subseteq X$ is open iff for each $x\in U$ there is an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\cap D\subseteq U$. Equivalently, $D$ is a dense open subset of $X$ whose relative topology is its usual one, and each $x\in ... 0 Here I am trying to give an intuitive idea...Let see how far I can make it clear... let$U,V$be two chart which cover these two spaces,$U\cup V=X$and$U,V$is homeomorphic to open disc$\mathbb{D^2}$...now since X is compact so boundary of$U$should properly contained in$V$similarly boundary of$V$contained inside$U$...now if I take a point$u$out ... 1 It seems the following. I shall call such spaces good, as a working term. Counterexamples. One of the simplest examples of not a good space is the square of the Sorgenfrey arrow, that is the real line endowed with the topology generated by the base$\{[a,b):a<b\}$. It is a separable space containig a closed discrete set$\{(x,-x) : x\in\Bbb R\}$of ... 3 Hint: take your favourite non-Hausdorff topological space$X$, and let$S$consist of two points that do not have disjoint neighbourhoods. 0 Since$A$is a zero dimensional Hausdorff space (I assume you mean a zero dimensional manifold, which means that each point has an open neighborhood homeomorphic to$\mathbb{R}^{0}$, i.e. the one point set) it is comprised entirely of isolated points. So, given any function$f:A\to X$and any open set$U\subseteq X$, we have$f^{-1}(U) = \bigcup_{a\in ...

1

Since $A$ and $B$ are compact, their product $A\times B$ is compact. Hence the function $d:A\times B\rightarrow \mathbb{R}$ has an element $x\in A\times B$ such that $d(x)$ is the infimum. This proves the theorem. For a counterexample, take $A=\{n+2^{-n}:n\in\mathbb{Z}^+\}$ and $B=\mathbb{Z}^+$. Note that $A\cap B=\emptyset$ but $\lim 2^{-n}=0$, hence ...

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The upper limit topology is finer. Notice that in the upper limit topology,\begin{equation*} (a,b) =\bigcup_{n\in\mathbb{N}}\left(a,b-\frac{1}{n}\right]. \end{equation*} Hence every open interval in the standard topology is also open in the upper limit topology, but something like $(c,d]$ is not open in the standard topology. I hope this makes sense.

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Hint What is $f([1, \infty))$? What is $f((\frac1{2\pi + \epsilon}, 1))$?

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This theorem is actually true generally. Theorem: If $S$ is a subset of a topological space $X$, then a set $U\subseteq S$ which is open in $S$ is open in $X$ iff $U\cap\partial S=\emptyset$. Proof: Suppose that $U\cap\partial S=\emptyset$, and let $x\in U$. Then $x$ is an interior point of $S$ because it is in $S$ and not on the boundary, so there is a ...

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Letting $\mathcal{F}$ denote the family of all nonprincipal ultrafilters on $\omega$, set $X = \omega \cup \mathcal{F}$ with the topology obtained by taking each point of $\omega$ to be isolated; and for each $p \in \mathcal{F}$ taking the basic open neighbourhoods of $p$ to be of the form $A \cup \{ p \}$ where $A \in p$. (This is called the strong ...

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your prove is not correct at the part "hence (X x X) - A = Union of all V(y)'s such that y doesn't belong to A.". This question can be solved directly by using the fact that every points have neighbourhood $U_1 \times U_2$ with$U_1,U_2$ is open subset. try to use thit :)

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Suppose $(a_n)_n$ is not bounded, then for every $n \in \mathbb{N}$, we find $m \in \mathbb{N}$ with $a_m \geq n$. Let $x^{(n)} \in \mathbb{R}^\omega$ be defined by $x^{(n)}_{m'} = 0$ for $m' \neq m$ and $x^{(n)}_{m} = a_m^{-1}$. Then $x^{(n)}$ converges with respect to $\tilde\rho$ to $0 \in \mathbb{R}^\omega$, the sequence consisting only of zeroes. But ...

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Yes you are right. Let's suppose $b_i=0$. And for if the map is continuous it mis be bounded above, since we're it not, we can find $a_{k_i}$ such that $a_{k_i}>i, i=1, 2, 3,\cdots$, then the sequence $(0,\cdots,1/1,\cdots,1/2,\cdots,1/3,\cdots),(0,\cdots,0,\cdots,1/2,\cdots,1/3,\cdots),(0,\cdots,0,\cdots,0,\cdots,1/3,\cdots),\cdots$ which converges to ...

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I am answering your first question. Executive summary: there's a (particularly nice type of) homotopy equivalence between three-space with axes removed and the cube graph. Claim 1: Let $V = \{(x,0,0)\} \cup \{(0,y,0)\} \cup \{(0,0,z)\}$ (three coordinate axes) and let $W = \{(\pm 1,0,0)\} \cup \{(0,\pm 1,0)\} \cup \{(0,0,\pm 1)\} \subset V$ (six ...

2

Assume that there exists an $n$ such that the implication $m\mid n\implies f (m)\mid f (n)$ does not hold. Then we find an $m$ such that $m\mid n$ but $f (m)\not\mid f (n)$. Now choose $U\in\tau$ to contain only the divisors of $f (n)$. Then $n\in f^{-1}(U)$ but $m\notin f^{-1}(U)$, so $f^{-1}(U)\notin\tau$, completing the proof of the desired ...

2

Suppose $f$ is continuous and $m|n$. Let $U$ be the set of $a$ such that $a|f(n)$. Then $U$ is open and $f(n)\in U$. Thus $n\in f^{-1}(U)$ which is also open. Thus since $m|n$, $m\in f^{-1}(U)$. Thus $f(m)\in U$. Thus $f(m)|f(n)$.

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We have 3 way to prove it,way one:definition of derivation .way two: is mean theorem for sub interval,way three is definition of concave function (this way is hard).we prove from way two :because $\varphi$ is concave then $\varphi^\prime$ is descending ,and apply mean theorem for intervals [0,a] and [b,b+a] assume $0<a<b$ ,for example the function ...

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"Naturally identified" leaves enough wiggle room to identify an infinite subset $X$ of $\mathbb N$ with the sequence in $\mathcal N$ that enumerates the elements of $X$ in increasing order. The set of all these increasing sequnces is a closed subset of $\mathcal N$.

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Let $Q$ be the space of non-negative rational numbers, $N$ the natural numbers including $0$, and $q:Q\to Y=Q/N$ the quotient map identifying $N$ to a single point. Note that $Y$ is a Hausdorff normal space since $Q$ is such a space and $q$ is a closed map. Now look at $Y\times Q$. This space is Hausdorff. We will show that it's not compactly generated. Let ...

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HINT: Let $\Phi$ be the set of all continuous functions from $Y$ to $[0,1]$. For each $\varphi\in\Phi$ let $I_\varphi$ be a copy of $[0,1]$, and define $$h:Y\to\prod_{\varphi\in\Phi}I_\varphi:y\mapsto\langle \varphi(y):\varphi\in\Phi\rangle\;.$$ Show that $h$ is a homeomorphism of $Y$ onto $h[Y]$. Now you can let $Z=\prod_{\varphi\in\Phi}I_\varphi$ or, ...

1

We show that $g$ is homotopic to $g*e$, by defining $H:[0,1]\times[0,1]\to X$ by$$H(x,t)=\left\{\begin{array}{rl}g((t+1)x)&0\leq x\leq\frac{1}{t+1}\\g(1)&\frac{1}{t+1}\leq x\leq1\end{array}\right..$$It is easy to define a similar homotopy between $g$ and $e*g$. Then, the two homotopies can be joined to one homotopy.

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With GEdgar's hint: (i) Show that a Cauchy sequence has a subsequence with the $r^{-i}$ property. Let $(x_n)$ be a Cauchy sequence, then for $i=1,2,3,\ldots$ there exists some minimal $N_i$ such that $d(x_j, x_k) < r^{-i}$ for $j,k \ge N_i$, set $n_{i} := N_i$ (by the minimality of the $N_i$ we have $N_1 \le N_2 \le N_3 \le \ldots$). Then the ...

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The only point that you get close to but not reach is $(0,0)$ which is the limit as $t \to - \infty$. So the closure is just the image plus this one point.

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$\mathbb Q$ is not a discrete space because points are not open. If you intersect an open set in $\mathbb R$ with $\mathbb Q$ you can never get a single point. Thus $\mathbb Q$ is not discrete.

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Recall that the $n$-dimensional volume of a ball of radius $r$ is $\pi_nr^n$ with $$\tag1 \pi_n=\frac{\pi^{n/2}}{\Gamma(\frac n2+1)}=\begin{cases}\frac{\pi^k}{k!}&\text{if n=2k}\\ \frac{k!2^n\pi^k}{n!}&\text{if n=2k+1}\end{cases}$$ (where $\pi_1=2$, $\pi_2=\pi$, $\pi_3=\frac43\pi$, etc.). Note that $\pi_n\to 0$ as $n\to \infty$. More ...

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$\newcommand{\Reals}{\mathbf{R}}$Let $f:(-1, 1) \to \Reals$ be your favorite smooth, increasing bijection, such as $$f(x) = \tanh^{-1} x \quad\text{or}\quad f(x) = \frac{x}{1 - x^{2}} \quad\text{or}\quad f(x) = \tan \tfrac{\pi}{2} x.$$ The mapping $$\phi(x_{1}, \dots, x_{n}) = \frac{\bigl(f(x_{1}), \dots, f(x_{n})\bigr)}{\sqrt{1 + f(x_{1})^{2} + ... 0 Take X=Y=S^1 (with elements angles \operatorname{mod} 2\pi) and f\colon \theta\mapsto 2\theta the usual 2-fold covering. Let V_1=(0,\pi) and V_2=(\frac{\pi}{2},\frac{3\pi}{2}) such that V_1\cap V_2=(\frac{\pi}{2},\pi)\neq\varnothing. Now f|_{V_1} and f|_{V_2} are both homeomorphisms onto their images U_1=(0,2\pi), U_2=(\pi,3\pi), ... 0 Of all the properties of a distance, the one requiring some work is the triangle inequality. Before tackling it, note the following: in (3) take \alpha = \frac 1 2, x_2=0, x_1=x. Then \phi (\frac 1 2 x) \geq \frac 1 2 \phi (x) (using also (1)). Since \phi is increasing and \rho satisfies the triangle inequality , \phi \circ \rho (x,y) = \phi (\rho ... 0 I proof the a) part, with this help b) can still be done as an exercise. Assume f is continuous in x \in X with respect to \bar{\rho}. Take \epsilon>0 we have to show:$$\exists \delta>0: \rho(x,y)<\delta \Rightarrow \sigma(f(x),f(y))< \epsilon$$We know:$$\exists \eta>0: \bar{\rho}(x,y)<\eta \Rightarrow ...

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1/ It seems the following. Yes, both the necessity and the sufficiency can be easily checked. Let $\{x_n\}$ be a sequence of vectors of the space $V$. Necessity. Assume that the sequence $\{x_n\}$ converges to a vector $x\in V$, $k\in\Bbb N$ be an arbitrary index, and $\varepsilon>0$ be arbitrary. Choose a number $\delta>0$ such that ...

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A closed orientable surface embeds into $\Bbb R^3$; a closed non-orientable surface does not (codimension 1 connected closed manifolds in $\Bbb R^n$ separate it into two pieces; one is compact, with boundary the closed manifold; the boundary of an oriented manifold is oriented). Because $X \# Y$ is oriented iff $X$ and $Y$ are, $X \# X$ embeds into $\Bbb ... 2 Hints:$X$is the graph of a continuous function$f: [0,\infty) \to [0,\infty)$(what is it?). Continuous functions map connected sets to connected sets. 1 I'm going to construct a homeomorphism between$(0,\infty)$and$S^1 - \{(0,1)\}$which can easily be extended to the compactification$(0,\infty)\cup \{p\}$and$S^1$. Let$f:(0,\infty) \to (-1,1)$be the map$f(x) = \frac{2}{x+1} -1$.$f$is clearly injective, continuous and if$r \in (-1,1)$then$x = \frac{2}{r+1}-1 \in (0,\infty)$will map to$r$. ... 1 A deformation retract of a space$X$to a subspace$A$is a continuous map$F : X \times [0, 1] \to X$such that$F(x, 0) = x$,$F(x, 1) \in A$and$F(a, t) = a$for all$a \in A$,$x \in X$and$t \in [0, 1]$. To prove that the moebius strip deformation retracts onto the center circle, consider the fundamental square$[0, 1] \times [0, 1]$with the two ... 0 Yes, just “pulling the sides in” works and it’s not hard to describe explicitly. Parameterize the planar representation$P$of the band below, where$-\pi\le x \le \pi$and$-1\le y\le1$, and where$e$(which you need not give explicitly) is any embedding of$P$into$\mathbb R^3$that’s one-to-one except two-to-one on the left and right sides ... 1 It is wellknown that interval$[-\pi,\pi]$is compact and function$f:[-\pi,\pi]\rightarrow\mathbb R^2$prescribed by$x\mapsto\langle\sin 2x,\sin x\rangle$is continuous. Consequently the image of this function is a compact subspace of$\mathbb R^2$. Moreover it coincides with the image of restriction$f\upharpoonleft\left(-\pi,\pi\right)$since ... 1 HINTS:$A$is closed so$A^c$is open...so there exists$\delta >0$s.t$B(x_0, \delta) \cap A = \phi$(since$x_0$is a interior point)...consider$U= B(x_0,\delta /2)$and$V= \cup_{a \in A} B(a,\delta/2)$since arbitraty union of open set is open... 3 No: an infinite space with the cofinite topology is compact and$T_1$but not Hausdorff. 0 Take the rectangles$[n,n+1]\times[-\varepsilon/2^{|n|},\varepsilon/2^{|n|}]$for$n\in\mathbb Z$and fixed$\varepsilon>0$. Then their area is$2\varepsilon/2^n$, and$\mathbb R\times\{0\}$is contained in the union of all these rectangles. Summing up the areas gives$6\varepsilon$, so the (two)-dimensional Lebesgue measure of$\mathbb R\times\{0\}$must ... 1 Let$\lambda$denote the Lebesgue measure. Then since$\mathbb{R}$and$\{ 0\}$are measurable sets, if$\mu=\lambda \times \lambda$is the product measure $$\mu(\mathbb{R} \times \{0\})= \lambda(\mathbb{R})\cdot\lambda(\{0\})=\infty \cdot 0 = 0$$ 6 $$m(\mathbb{R}\times\{0\})=m\left(\bigcup_{n=1}^\infty[-n,n]\times\{0\}\right)\leq\sum_{n=1}^\infty m([-n,n]\times \{0\})=\sum_{n=1}^\infty (2n)\cdot 0=0.$$ Intuitive explanation: A line has no area. 0 Your argument uses the axiom of countable choice, which says that every countable family of non-empty sets has a choice function. This question (one of those to which you linked) shows that some amount of choice is needed. The axiom of choice is usually not mentioned in such arguments simply because it’s generally taken for granted in doing topology. 1 Yes, it is. Notice that$baN=ab(b^{-1}a^{-1}ba)N=abNabaN=a(baN)=a(abN)=(a^2)bN=bNbabN=aN$for the same reason. Therefore, no:$G/N$is indeed the Klein 4-group. Added Notice that if a normal subgroup contains$abab$, then it must contain$a^{-1}(abab)a=baba$... 1 Yes. The free product of more than two *non-trivial_ groups is always an infinite group. About$\;N=\langle\,g_1g_2g_1^{-1}g_2^{-1}\,\rangle^G=\langle\,(g_1g_2)^2\,\rangle^G\;$, but why would this equal$\;1\;$? It doesn't, as then$\;G\;$would be finite: you do not have commutativity between elements of the two free factors. 2$N$seems to be the commutator subgroup of$G$, and so the quotient should just be the Klein-four group. 0 This statment is equivalent to the statement$\forall x \in X. \exists U_x. \forall g \in G. g \neq e \Rightarrow g\cdot U_x \cap \cdot U_x = \emptyset$, by translating your definition by$g^{-1}$. Let$X = \mathbb{R}$, and$G = (\mathbb R, +)$be the additive group acting on$X$by translation. This is clearly a free action. Now for$x= 0\in X$, you would ... 3 Take$X = \mathbb N \cup \{ - \infty , + \infty \}$with the topology where each point of$\mathbb N$is isolated, each neighborhood of$- \infty$and$+ \infty$is a cofinite subset of$X$(containing the respective point). Note that for each$n \in \mathbb N$the singleton$\{ n \}$is clopen in$X$. As$- \infty , + \infty\$ cannot be separated by ...

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