New answers tagged

1

Since it is suggested that connectedness be explicitly used, I might phrase it like this: The image of the connected domain of the function $\mathbf z \mapsto \phi(\mathbf z) - \phi(-\mathbf z)$ must be a connected subset of $\mathbb R$. If it is not everywhere $0$, then for any point $\mathbf z_0$ where it is not $0$, the function changes signs as $\mathbf ...


3

Can you use continuity and the Intermediate Value Theorem for a Connected domain to show that the function $$ \phi(z)-\phi(-z)=0$$ for some $z$? If so, you are done.


0

For your first question the answer is positive. It is not very hard to prove the completeness of $C[0,1]$ (or actually any $C(K)$ where $K$ is compact), you should first try doing it on your own. For the second question, under the usual metric (which is just what you provided in your post as $d(f,g)$), if a sequence $\{f_n\}$ converges in $C[0,1]$ then it ...


1

Hint for an explicit bound in (1): $$\forall x\in\Bbb R:\ 1-\frac{x^2}2\le\cos x\le 1.$$ (2): Yes, because $|b\cos(x/n)-b| = |b|\,|\cos(x/n)-1|$. (3): Yes.


1

For the first question, note that $ \cos(x)$ is positive and decreasing for $x \in[0, \frac{\pi}{2}]$. So $f_n(x)$ is also positive and decreasing on $[0,1]$, thus we have $$d(1,f_n) = \sup_{x \in[0,1]} 1 - \cos \left( \frac{x}{n} \right) = 1 - \cos \left( \frac{1}{n} \right)$$ So you are left to show that $\cos \left( \frac{1}{n} \right) \to 1$ as $n \to ...


0

Definitely it is meant uniform convegence since a uniform limit of a sequence of continuous functions is again continuous while a pointwise limit of continuous functions is not continuous in general. The convergence is always related to your topological structure of your space, and in your case, the topological structure is just a topology induced by metric. ...


5

This is a global condition - that is, it is both necessary and sufficient to have your condition be true for all $x,\delta$. You need: (Condition 1): Given any $x\neq y$ and any $\epsilon>0$ that there is some $z$ so that $d(y,z)<\epsilon$ and$d(x,z)<d(x,y)$. That is, every neighborhood of $y\neq x$ has a point closer to $x$ than $y$ is. ...


0

An example of a meaningful, sufficient condition for this is that $X$ is a length space. This is certainly not a necessary condition: for example, a dense subspace of a length space also has this property (more generally, the property is inherited by dense subspaces; it is also inherited by open subspaces). A simple necessary condition is that for each $x$, ...


1

Suppose you've got a set, $X$. If you equip $X$ with a metric $d$, now the pair $(X,d)$ is a metric space. This metric generates a topology on $X$. You consider the collection $\mathcal{B}$ of sets of the form $$ B(x,r):= \{y \in X:d(x,y) < r \}. $$ Now the collection $\mathcal{B}$ is not a topology in and of itself, but lets you build one by taking ...


1

Except for the trivial case of a metric space with only one element, there is always at least one topology on a metric space that is not the same as the metric topology, namely the discrete topology in which only the empty set and the whole space are open. An example of a very interesting topology on $\Bbb{R}$ that is not the metric topology is the lower ...


0

The easiest way of seeing this is to remember that compactness is equivalent to sequential compactness in metric spaces. Now, as $[a,b]$ is not a complete metric space, find a Cauchy sequence without a limit inside $[a,b]$ (say a sequence whose limit would be $a+\frac{\sqrt2(b-a)}2$). Now you have a sequence in the interval without a convergent subsequence, ...


0

Since $\left|p^n\right|_p = d_p\left(p^n,\,0\right)=p^{-n}$ for any $n\in\mathbb{Z}$, $\lim_{n\to\infty}\left|p^n\right|_p = 0$. Thus $\lim_{n\to\infty}\left|p^n - L\right|_p = 0$ for $L = 0$, which is therefore the limit of the sequence.


2

In a metric space $(X,d)$, you define the metric interval between $x,y\in X$ as $$I(x,y) = \{z \in X \mid d(x,y) = d(x,z) + d(z,y)\}$$ Note that this metric interval is always closed (think sequences and continuity of $d$). With this notion, you can start to define and study things like convex and hyperconvex metric spaces, etc, despite not having an ...


1

Suppose $x_n$ is a sequence in $Z(f)$ such that $x_n\to x\in X.$ We want to show $x \in Z(f).$ By continuity, $f(x_n) \to f(x).$ But $f(x_n) = 0$ for all $n,$ so …


2

The theorem in your book says that "if two points in $Y$ are to close, then their preimages are to close" (remember: closeness is measurable by mean of open sets in general topology). It has a good corollary: $f$ is continuous iff $f^{-1}(V)$ is closed for every closed $V$. Now, i'll suppose you are working with the usual topology. In your case, note that ...


4

It is not a matter of "thinking", "considering" or "debating". Your professor perhaps has given a definition of open ball. Or, at least, he must have assumed some definition. That definition should specify if the radius of the ball must be a positive number or null radii are allowed. From my experience, most books that include a definition of open ball say ...


4

Start with the fact that for any homeomorphism of topological spaces $f:X \to Y$ and for any $x \in X$ and $y=f(x) \in Y$, the restricted function $f : X-\{x\} \to Y-\{y\}$ is a homeomorphism using the subspace topologies. So for your problem it suffices to show that if $X=Y=B_1[0]$ and $x \in S^n$ and $y \in B_1[0]-S^n$ then $B_1[0] - \{x\}$ is not ...


3

One possibility is to use the Invariance of Domain theorem (which in turn relies on non-trivial topology). Let $h\colon B_1\to B_1$ be a homeomorphism. Here $B_1$ is the closed unit ball in $\mathbb{R}^n$. Let $\mathring{B}_1$ denote the open unit ball. The restriction of $h$ to $\mathring{B}_1$ is an injective continuous map of a domain of $\mathbb{R}^n$ ...


1

Hint: your space is complete, so Cauchy $\implies$ convergent. Now, what happens with the pointwise limit?


5

I understand the bi-invariance as $$ d(ax,ay)=d(x,y)=d(xa,ya) $$ for any $a,x,y\in G$. Then $$ d(x,y)=d(1,x^{-1}y)=d(y^{-1},x^{-1})=d(x^{-1},y^{-1}). $$ The last step is the symmetry of $d$.


1

I will answer for $n = 2$. Then you can generalize to higher values of $n$. The argument is similar. Given two points $(x_1,y_1)$ and $(x_2,y_2)$, say I know $M = \phi^{*}(x,y) = |x_1 - y_1| + |x_2 - y_2|$. How large can $\phi^{+}(x,y) = \max(|x_1 - y_1|, |x_2 - y_2|)$ be? The sum of the two nonnegative numbers $|x_1 - y_1|$ and $|x_2 - y_2|$ is $M$. ...


1

Still another possibility is to rewrite your set: $$ A=f^{-1}(0,1),$$ where $$ f(\boldsymbol{x})=\max_{j\ge 1} \lvert x_j\rvert,\qquad \boldsymbol{x}\in\ell^2$$ is a function $$ f\colon \ell^2\to [0, \infty).$$ If you show that $f$ is continuous then $A$ is open, being the preimage of an open set. But $f$ is actually a norm, and so it is continuous if and ...


0

No, you do not need a metric space for "the application of the Algebra of Events"--as your nomenclature suggests, all you need is something like an algebra, technically a $\sigma-algebra$ in most models. Assigning (possibly arbitrary) notion of distance to a random variable for the purposes of regression is separate from the ability to ask questions about ...


0

$l_p$ metric on $\mathbb R^m$: $$ d_p(x-y) = \left(\sum_{i=1}^m \big|x_i-y_i\big|^p\right)^{1/p} $$ $1\le p < \infty$. I will let you do the $l_\infty$ metric.


1

(some changes in order to improve precision) About your last interrogation. Any normed vector space defines naturally a metric space by the relationship $d(x,y)=\|x-y\|_p$. Thus, indeed, any $\|\cdot\|_p$ norm induces naturally an $\ell_p$ "metrics" (synonym : "$\ell_p$ distance"). A point of vocabulary about the words "metrics" vs. "distance". "Metrics" ...


1

Another approach. Let $$U=\{f\in C[0,1]:\mbox{ there is $K$ such that}|f(x)-f(x_0)|\le K|x-x_0|\,\,\forall x\in[0,1]\}$$ Note that $A\subset U$. On the other hand, if $f,g\in U$ and $a$ is scalar: $\begin{eqnarray}|(f+ag)(x)-(f+ag)(x_0)|&=&|f(x)+ag(x)-f(x_0)-ag(x_0)|\\ &\le&|f(x)-f(x_0)|+|ag(x)-ag(x_0)|\\ ...


1

Let $f$ be a function in $A$. $f$ belonging to the interior of $A$ means that there is a "small ball" in $A$ wrapping it. So to disprove it you need to find a way to show that this kind of "small ball"s does not exist. The simplest and straightest way is to find a sequence $(f_n)_{n\in\mathbb{N}} \notin A$ that converges to $f$ in this metric. For example ...


0

The limit of a finite sum is the finite sum of the limits, and the limit of an absolute value is the absolute value of the limit.


3

Let $U = (-1,1)$, $V_x = (-1,1)$ for $x = 0$ and $V_x = (-\frac{1}{2}, \frac{1}{2})$ otherwise. There is obviously no $\varepsilon$-Ball around $(0, \frac{3}{4})$.


2

Not sure why you use $a$ both for elements in $\mathbb R$ and in $X$. Note by triangle inequality, $$d(x, a)\le d(x, y)+ d(y, a)\Rightarrow d(a, x) - d(a, y)\le d(x, y)$$ Interchanging the role of $x, y$ and use $f(x) = d(a, x)$ we have $$\tag{1}|f(x) - f(y)|\le d(x, y).$$ This inequality is sufficient for us to condlude that $f$ is continuous. Let $U$ ...


2

The open character of A and B can be deduced from the definition of open set. Let $a\in A$. Supposing $b\in B$^, $(a,b)\in A\times B$. As $A\times B$ is open, there exists $B((a,b),r)\subset A\times B$. Then if $a'\in B(a,r)$, $d_X(a,a')=d_X(a,a')+d_Y(b,b')=d_1((a,b),(a',b))<r$, so $(a,b)\in A\times B$, and thus $a'\in A$. That is, A is open. An analogous ...


1

$A\times B$ is open implies for every $(a,b)\in A\times B$, there exists a ball $B((a,b),r)\subset A\times B, r>0$. $(x,y)\in B((a,b),r)$ i.e $d_X(x,a)+d_Y(y,b)<r$. Let $B(a,r)=\{x\in X:d_X(x,a)<r\}$, we have $B(a,r)\times b\subset B((a,b),r)\subset A\times B$. So $B(a,r)\subset A$ and $A$ is open. Same argument with $B$.


0

An elaboration on Surb's comment For clarity, I'll use $d_\infty((x_1,x_2),(y_1,y_2)) = \max\{|x_1 - y_1| , |x_2 - y_2|\}$, and $d_2((x_1,x_2),(y_1,y_2)) = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2}$. Observe that for any $(x_1,x_2),(y_1,y_2) \in \Bbb R^2$, $$d_\infty((x_1,x_2),(y_1,y_2)) \le d_2((x_1,x_2),(y_1,y_2)) \le \sqrt 2 d_\infty((x_1,x_2),(y_1,y_2)).$$ ...


4

The triangle inequality implies that $|d(y,x) - d(y',x)| \leq d(y,y').$ It follows that $d(-,x) : X \to \mathbb{R} $ is continuous (in fact, Lipschitz). Hence, the preimage under the open subset $]r,\infty[ \subseteq \mathbb{R}$ is open, and this is your set.


3

Let $y\in X$ such that $d(y,x)>r$, and choose $\epsilon$ such that $0 < \epsilon < d(x,y) - r$. We claim that $$B(y,\epsilon):= \{z\in X|d(y,z)<\epsilon\} \subseteq \{y\in X|d(y,x)>r\},$$ so the open ball $B(y,\epsilon)$ is contained in any arbitrary point $y$ in the given subset of $X$. Let $z \in B(y,\epsilon)$, then $d(y,z)<\epsilon ...


1

A necessary and sufficient condition on a metric space $(X,d)$ for $d$ to come from some pre-metric is that $(X,d)$ be isometric to a subspace of the metric space $\mathbb{R}$ with the ordinary distance. We may assume $X \ne \varnothing$. (When $X = \varnothing$, both conditions are true.) Let $f$ be a pre-metric on $X$, and fix an "origin" $a \in X$. ...


1

Your proof is good, but too lengthy. Suppose $E$ is not bounded. For each positive integer $n$, there is $x_n\in E$ such that $\|x_n\|>n$. I claim that $S=\{x_n:n>0\}$ is infinite. If it is finite, then $M=\max\{\|x_n\|:n>0\}$ exists; but if $n$ is the least integer greater than $M$, we have $\|x_n\|>n>M$: a contradiction. The set $S$ cannot ...


0

I didn't read your proof ( I apologize), but here's my attempt: Do this by contrapositive. First suppose $E$ is not closed. Then $E^c$ is not open, thus $\exists$ $x \in E^c$ such that $\forall$ $\epsilon>0$, $V_\epsilon(x)\nsubseteq E^c$, where $V_\epsilon(x)$ denotes an open neighborhood of radius $\epsilon$ centered at $x$. Then $\forall$ $n \in ...


0

Let us assume that bd($B_\epsilon(x)$) and bd($V$) are non-empty. Given, bd($B_\epsilon(x)$) = bd($V$). We have, $V \subseteq B_\epsilon(x)$. Assume, if possible, that $x \notin V$ then $d(x,V)>0 \implies $ there exist $a \in bd(V)$ such that $d(x,a)=min_{z \in \overline{V}}{d(x,z)}>0$. Now, there exist an open ball around $a$ which lies entirely in ...


1

Hint,for ${x_n},{y_n}$ belong to $l^2$,c=sup|$x_n|<1,$ let $\epsilon$=1-c,if||x-y||<$\epsilon$,then,|$x_n-y_n$|<$\epsilon$,any n.$


1

The key point is that, since $\forall x\in\ell^2,\ \lim_{n\to\infty}x_n=0$, it must hold $$\xi\in A\implies \exists\, \alpha_\xi>0\ \ \forall n,\ \lvert \xi_n\rvert<1-\alpha_\xi$$ So, $\varepsilon=\alpha_\xi$ works, since it holds, in general, $$\forall n,\ \lvert x_n-y_n\rvert\le d_2(x,y)$$


1

In higher dimensions this is very difficult and, I think, there is no simple characterization along the lines of the ones in low dimensions. The main difficulty comes from characterization of compact manifolds among, say, compact metrizable topological spaces. The characterizations of spheres in dimensions 1 and 2 are (implicitly) based on the fact that in ...


0

I would say that your proof is better. I would assume that he missed a detail, and left out a proof that if it holds for $E$ then it holds for $\overline{E}$. Certainly, if I were grading a course I would mark his proof as incomplete - even in a course not for first or second years. Especially since his book is a standard introductory text for first and ...


4

Yes. Ignoring questions of measurability: Suppose $f:[0,1]\to X$ and $\int_0^1||f(t)||_X^2\,dt<\infty$. Say $X_n$ is an increasing sequence of finite-dimensional subspaces of $X$ so that $\bigcup X_n$ is dense in $X$. For each $n$ choose $f_n:[0,1]\to X_n$ such that, say, $$||f_n(t)-f(t)||_X\le 2 d(f(t),X_n)$$for all $t\in [0,1]$. Note that ...


0

As $X \setminus E$ is open, and $a$ is in it, there is some $r>0$ such that $B(a,r) \subseteq X \setminus E$, or $B(a,r) \cap E = \emptyset$. So for all $x \in E$, $d(x,a) \ge r$ (or otherwise it would be a point in the ball and in $E$ at the same time, which cannot be). So the $\inf_{x \in E} d(x,a) \ge r > 0$, as $r$ is a lowerbound for the ...


1

If there are points $x_n\in E$ arbitrarily close to $a$ then there is a subsequence of the $x_n$ that converges to $a$ so $a$ is in the closure of $E.$


0

Supose $(a_n)_{n\in N}$ is a sequence of points of $\bar A$ converging to $a,$ where $a\not \in \{a_n:n\in N\}.$ For each $n\in N$ let $b_n\in A\cap B_d(a_n,d(a,a_n)).$ Then $(b_n)_{n\in N}$ is a sequence of points of $A,$ and $a\not \in \{b_n :n\in N\}.$ We have $\lim_{n\to \infty}d(a,a_n)=0,$ and for each $n\in N$ we have $d(a,b_n)\leq ...


3

In 2-dimensions you might want the Kline sphere characterization theorem.


2

If $(0)$ is the original definition, $(0)\implies(1):\quad f^{-1}(F)=f^{-1}(F^c)^c$ $(1)\implies(2):\quad f^{-1}(\overline{f(A)})$ is closed and $A\subset f^{-1}(f(A))\subset f^{-1}(\overline{f(A)})$. $(2)\implies(3):\quad$ Replace $A$ with $f^{-1}(B)$. $(3)\implies(1):\quad$ Replace $B$ with $F$ and remember that $\bar{F}=F$. ...


1

Let $x$ be a limit point of $\overline A$ that isn't a limit point of $A$. Then there exists an $\epsilon > 0$ such $B_{\epsilon}(x)$ will contains no point of A. But as $x$ is a limit point $\overline A$ it must contain a point $y \in \overline A/A $ so $y$ is a limit point of $A$. Let $0< \epsilon_2 = \epsilon - d(x,y)$. $B_{\epsilon_2}(y)$ must ...



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