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0

The answer is positive. Note that the closure is a (compact) convex set, and in particularly star-shaped. (We call a set $U⊆\mathbb{R}^n$ star-shaped if there is a point $x∈U$ such that for all lines $L$ through $x$, $L∩U$ is a connected open line segment). Now we use the star-shaped structure. My answer is essentially taken from here. Translating $U$ if ...


1

In symbolic dynamics people use similar metrics for spaces of sequences and the motivation for the formulas is always the following: "the longer central block on which two sequences coincide, the closer they are to each other". I think in your undefined case sequences $x$ and $y$ "coincide on block of zero length", so it's tempting to say that you assign ...


0

I think the result is true because mapping connected sets to connected sets for metric spaces implies continuity for bijections. See my book draft: http://www.geometry.org/tex/conc/dgstats.php, currently Section 28.2, page 869. Then if continuity of a bijection for such spaces implies continuity of the inverse, you get the result you want.


6

One could also argue in the following way: Assume we know that $(\mathbb{R},d(x,y))$ is a complete metric space, then the set of natural numbers $\mathbb{N}$ is a closed subset of $\mathbb{R}$, so it must hold that $(\mathbb{N},d(x,y))$ is also a complete metric space with respect to the same metric since closed subsets of complete spaces are complete too. ...


1

If $\{x_n\}_n$ is Cauchy in this space and given $1>\epsilon>0$, if we have $|x_m-x_n|<\epsilon$ then $x_n=x_m$, in particular $x_n$ converge and your limit is a natural number.


4

If $x_n$ is a Cauchy sequence, pick $\varepsilon=\frac12$, then for some $N$, it holds that every $n,m>N$ satisfy $|x_n-x_m|<\frac12$. What does that tell you?


5

Suppose that $X$ is a complete, countable, homogeneous metric space. Clearly $X=\bigcup_{x\in X}\{x\}$ is a countable union of closed sets. Every complete metric space is a Baire space, so the sets $\{x\}$ cannot all be nowhere dense. Thus, at least one of them must be isolated, and since $X$ is homogeneous, they are all isolated, and $X$ is discrete.


2

Suppose that $V$ is a vector space over $\mathbb R$, $\|\cdot\|$ is a norm on it, and $$U\equiv\{v\in V\,|\,\|v\|<1\}$$ is the open unit ball with respect to the norm. Fix $x\in V$ and $x\neq 0$. Claim 1: There exists some $t>0$ such that $tx\in U$. Proof: Since $x\neq 0$, one has $\|x\|>0$. Put $t\equiv 1/(2\|x\|)>0$. Then, ...


0

Norms are inspired from the Euclidean distance function and refer to a generalized class of metrics $d$ which for a normed linear space $V$, satisfy the properties: $d(a,b) = d(a-b,0) = d(0,b-a)~\forall~a,b \in V$ $d(\lambda u,0) = |\lambda| d(u,0)~\forall u \in V$ $d(a,b) \le d(a,c)+d(c,b)$ $d(a,b) \ge 0$ with equality $\iff a=b.$


3

As Hagen von Eitzen said in his comment, the two definitions are equivalent only as long as $X \neq \varnothing$. Other than that, the $(\Leftarrow)$ part is correct, but the $(\Rightarrow)$ part is not. Namely, when going from 4 to 5 you should use $d(p,q) \leq d(p,q_0) + d(q,q_0) < 2M_0$ instead of $d(p,q) < M_0$. From your question it isn't clear ...


0

No, $\sigma$ is not the average $\ell^2$ distance from $\mu$. The $\ell^2$ distance from $x$ to $\mu$ is $\sqrt{(x-\mu)^2} = |x-\mu|$ and the average of those is $$ \sum_{x\in X} p(x)|x-\mu|. $$ Rather $\sigma$ is the $\ell^2$ distance from the tuple of $x$ values to the tuple in which every component is $\mu$. A reason for the use of the mean squared ...


2

Hint. Use two ingredients to prove the completness of $\mathcal C^2([0,1])$: Ingredient one the space $\mathcal C([0,1]$ of continuous real functions defined on $[0,1]$ is complete for the $\Vert f \Vert =\sup\limits_{x \in [0,1]} \vert f(x) \vert$ norm. Ingredient two if a sequence of differentiable functions $(f_n)$ is such that $f_n^\prime$ converges ...


0

Put $\mathbf{v} = \mathbf{x} - \mathbf{y}$ Then $d(x,y) = \|\mathbf{x}-\mathbf{y}\| = \|\mathbf{v}\|$ and $\langle{\mathbf{v},\mathbf{v}}\rangle = \langle{\mathbf{x}-\mathbf{y},\mathbf{x}-\mathbf{y}}\rangle = \displaystyle{\sum\limits_i^n(x_i-y_i)^2}$ and so $\|v\| = \sqrt{\langle\mathbf{v},\mathbf{v}\rangle}$ implies the desired result.


1

$$ \sum_i (x_i + \lambda y_i)^2\geqslant 0 $$ $$ \sum_i x_i^2+\lambda x_iy_i +\lambda x_iy_i+\lambda^2y_i^2 \geqslant0 $$ Let $\lambda=-(\sum_jx_jy_j)/(\sum _jy_jy_j)$ then: $$\sum_i x_i^2-\frac{(\sum_ix_iy_i)(\sum_jx_jy_j)}{(\sum _jy_jy_j)}-\frac{(\sum_ix_iy_i)(\sum_jx_jy_j)}{(\sum _jy_jy_j)}+(\frac{\sum_jx_jy_j}{\sum _jy_j^2})^2 \sum_iy_i^2 \geqslant0$$ ...


0

Use the infinite pigeon-hole principle: If aninfinite set is presented as the union of finitely many subsets,at least one subset is infinite.So if a sequence p(n) converges to p, where each p(n) belongs to some F(j), then,for at least one j, there are infinitely many n for which p(n) belongs to F(j), so p belongs to this F(j).


0

The absolute value of $|a|$ is $a $ or $-a $, depending on whether $a $ is nonegative or not. In your case, the condition $n\geq m $ implies $f_m (t)\geq f_n (t) $, so $$|f_n (t)-f_m (t)|=-(f_n (t)-f_m (t))=f_m(t)-f_n (t). $$ For your second question, when you calculate your integrals you are forgetting the antiderivatives.


2

No, it’s not true that every closed set in $\Bbb R$ with the lower limit topology is a union of sets of the form $\Bbb R\setminus[a,b)$. For example, $\{0,1\}$ is a closed set, because its complement is the open set $$\bigcup_{x<0}[x,0)\cup\bigcup_{0<x<1}[x,1)\cup\bigcup_{x>1}[x,x+1)\;.$$ Let $\tau$ be the lower limit topology. It’s not hard to ...


0

Open sets are going to look like arbitrary unions of $[a,b)$. Closed sets are defined to be the complement of an open set. So if $U$ is open, $$U = \bigcup_{i \in I} {[a_{i}, b_{i})}.$$ Then, $$ X - U = X - \bigcup_{i \in I} {[a_{i},b_{i})} = \bigcap_{i \in I} {[a_{i},b_{i})^{C}}$$ So your closed sets are looking like intersections of the sets you mentioned ...


1

The integral (also known as tha $L^1$ metric) and the uniform metric cannot be defined on any space. The uniform metric is defined on sets of bounded, real (or complex) valued functions defined on a set $X$. If $f,g\colon X\to\mathbb{R}$ are bounded, then $$ d_\text{u}(f,g)=\max_{x\in X}|f(x)-g(x)|. $$ It measures the greates difference between the values of ...


1

I know from your other questions you like to see rigorous answers; so I thought I'd do the same on this one too. I did this a slightly longer way since I do not know which definition of closed you are using (so I used a standard definition of open) def. $U$ is open $\Leftrightarrow \forall x \in U\; \exists \varepsilon >0 : B(x, \varepsilon) \subset U$ ...


0

For example if $X= \{0, 1, 2\}$ with $d(x,y)$ defined as $d(x,y)=|x-y|$ then there are only $2^3=8$ subsets so at most 8 "open balls".


0

It is not clear to me what you are asking exactly. Given a ball $B=B(x;r)$ in $\Bbb R^n$ all the balls $B(x;\rho)$ with $\rho<r$ are inside $B$ and so in whatever open set $A$ the ball $B$ is a subset of. This is entirely trivial. A bit less trivial question would be if you can find infinitely many disjoint balls inside an open set $A\subseteq\Bbb R^n$, ...


2

It may be the case that these open balls are actually equal to each-other despite having different radii. For instance, if we equip $\mathbb{Z}$ with the usual metric, then $B_{1/2}(0) = B_{1/3}(0)$, for instance.


9

In any metric space, there are an infinite number of ways to write down balls with a given center. But some of the balls might actually be the same. For instance, in the "discrete metric" $d(x,y)=0$ if $x=y$ and $1$ otherwise, all balls $B_r(x)$ for $r \leq 1$ are the same (they are just $\{ x \}$) while all balls $B_r(x)$ for $r>1$ are also the same ...


2

$\int_0^1\vert f_n(x)-1\vert dx=\int_0^{1/n}\vert f_n(x)-1\vert dx+\int_{1/n}^1\vert f_n(x)-1\vert dx$ Since $f_n(x)=1$ for $x\geq\frac{1}{n}$, the second term is equal to $0$. You can also remove the absolute value in the first term since for $0\leq x\leq\frac{1}{n}$, $f_n(x)=nx\leq 1$.


4

I'd say so. A sum of non-negative reals is 0 iff every term was itself zero. This is easy to prove independently.


1

I find doing proofs this way (especially in analysis) very helpful in understanding the details of what is happening. Here is a complete proof for the forward direction; of course I would shorten in considerably if I were to tun it in as an assignment or just thinking through it for myself. $f: (X, d) \rightarrow (Y, \rho)$ is uniformly continuous $E ...


0

There's more than logical inference to be done in this step. I think that, in effect, you need the following lemma: For all real numbers $r$, if for every $p,q \in E$ we have $d(p,q) \le r$, then $\operatorname{diam}(E) \le r$. The proof of this lemma essentially involves looking at the definition of $\operatorname{diam}(E)$, which is defined as a ...


1

No. It is not true. Let $A$ be the unit ball centered in the origin, let $D$ be the points with rational coordinates. Let $U$ be the complement of $A$.


0

It is only true that $U\cap D$ and $A\cap D$ are non-empty. $A\cap U$ can happen to be the empty set. Why $U\cap D$ is nonempty: Because $U$ is open, then for each $u\in U$ there is a ball $B(u,r)$ with center $u$ and some positive radius $r>0$, such that $B(u,r)\subset U$. Now, because $D$ is dense in $\mathbb R^n$ it follows that for each ...


0

Take $V=C[0,1]$ with the uniform norm, $A$ is the subspace of all polynomials (convex), $D=C[0,1]\setminus A$ (dense). But $A\cap D=\emptyset$.


0

First, $U\cap D\cap V=U\cap D$ because $U\subset V$. Second, any non-empty open subset of $V$ intersects $D$, precisely because $D$ is dense in $V$ (this is a definition of dense subset). So yes, $U\cap V\cap D$ is non-empty. Note that we have only used the topology of $V$. There's no need of linearity, convexity, norm, etc.


0

Yes. Let $B_m$ be the open ball of radius $r_m$ around $(x_m,y_m)$. It suffices to consider the case $n=2$ and here to show that there exists a path from $(0,0)$ to $(1,0)$ (assuming wlog. that these points are in the set). The obvious attempt is the curve $\gamma\colon [0,1]\to\mathbb R^2$, $t\mapsto (t,0)$, but for each disk it intersects we need to take ...


0

HINT: Let $\theta$ any real number such that $\frac{\theta}{\pi}$ is irrational. The metric space $(e^{i n \theta})_{n\ge 0}$ is isometric to the subset $(e^{i n \theta})_{n\ge 1}$ by the map $z \mapsto e^{i\theta} \cdot z$


0

In some cases, easier way to prove that a function is bijective is to show that it is invertible. (See Inverse of a Function exists iff Function is bijective.) In your case you have $f \colon T \to K$ defined by $f(U)=U^c$. It is relatively easy to see that $g \colon K \to T$ defined by $g(C)=C^c$ is inverse to this function. Since $f^{-1}$ exists, the ...


1

In general, if the space we're looking at is being viewed as a self-contained space, i.e., not a subspace of another space, then the entire space is trivially open. We know that $(\mathbb{R},d)$ is open as a metric space. Now here's the subtlety: Consider say $M=[0,1]$. Then $M \subset \mathbb{R}$ is not an open set. That is, if $M$ is contained in the ...


2

The distance $\rho(x,A) = \inf\{\rho(x,a) \mid a \in A\}$ is usually known as the Hausdorff distance and also generalises nicely to a distance between two sets. Now, what should $\rho(x,\emptyset)$ be? Intuitively, $\rho(x,A)$ gives the smallest distance between $x$ and any element in $A$. But since there are no elements in $\emptyset$, it makes some sense ...


0

Although the definition of GH metric mentions embeddings into an arbitrary metric space $X$, we can always truncate $X$ to the union of the image of $A$ and the image of $B$. In other words, it suffices to consider semimetrics on the abstract disjoint union $A\sqcup B$ that are compatible with the given metrics on $A$ and $B$. (Semimetrics are allowed to ...


2

The answer to your question is yes, the closure of $D_n$ is $X$. This is almost immediate: since $D$ is the intersection of the sets $D_n$, we must have $D \subseteq D_n$ for every $n$. It is a standard fact that for every pair of sets $A,B$, if $A \subseteq B$ then $\overline{A} \subseteq \overline{B}$. (Note that $\overline{B}$ is a closed set that ...


1

Similarly to the other answers, we will need to compare the usual distance of $\mathbb{C}^n$ and $\delta$ (which, as you noted, does not induce a distance). Let $m=\inf\left\{\delta(z):\Vert z\Vert=1\right\}$ and $M=\sup\left\{\delta(z):\Vert z\Vert=1\right\}$. The condition on $\delta$ implies $$m\Vert z\Vert\leq\delta(z)\leq M\Vert z\Vert$$ for all $z\in ...


0

If $E$ is a vector space over a non-discrete valued field $F$, then you can define a norm over it this way: Choose a basis $\beta: I \rightarrow E$ for $E$. For $x = x_{i_1}.\beta(i_1) + ... + x_{i_n}.\beta(i_n) \in E$, set $N_{\beta}(x) = \max(|x_{i_1}|,...,|x_{i_n}|)$. $N_{\beta}$ is a norm. (The existence of a basis for any vector space can be deduced ...


0

This is too long for a comment, but not really an answer. You're talking about metric spaces Every inner product induces a norm and every norm induces a metric There are examples on each page. This picture pretty much sums up how they relate here Lastly this shows how to get a distance from an inner product You know have definitions to play with Here ...


5

We often find metrics on function spaces from norms, i.e., $d(f,g):=\|f-g\|$. As norms you can take for example $\|f\|_\infty :=\sup\{\,f(x)\mid x\in I\,\}$ if $I$ is a compact interval and we consider the space of continuous functions on $I$. Or $\|f\|_2:=\int |f(x)|^2\,\mathrm dx$ for square-integrabvle functions. And many more.


5

This determines a norm $\|\_\|_P$ iff $P$ is positive definite, then it naturally defines a metric by $d(x,y):=\|y-x\|_P$. Else the above $d$ function would not be defined or would fail to be a metric, as e.g. there could be $x\ne 0$ with $x^TPx\le0$. All in all, what it basically says is that the quadratic form $x\mapsto x^TPx$ can be viewed as a ...


2

There is a Lipschitz function $\phi:[0,\infty)\to[0,1]$, with Lipschitz constant exactly $1/\epsilon$, such that $\phi(0)=1$ and $\phi(t)=0$ for $t\ge\epsilon$. Let $f(x)=\phi(d(x,A))$.


1

You have to interpret open as relatively open in $X$, since $X$ need not be open in the space. Suppose that $X=A\cup B$, where $A$ and $B$ are non-empty and separated. Then $A\cap\operatorname{cl}B=\varnothing$, so $$B\subseteq\operatorname{cl}_XB=X\cap\operatorname{cl}B\subseteq X\setminus A=B\;,$$ and it follows that $\operatorname{cl}_XB=B$. Thus, $B$ ...


4

Aside from coincidences, there are only $3!=6$ possible linear ordering among $3$ objects. So your table only needs $6$ rows to deal with all non-trivial cases. If you have $8$ rows, it is reasonable to expect that $8-6=2$ of the rows will be somehow degenerate. More specifically, if $y-x\geq0$ and $z-x\leq 0$ and $y-z\leq0$, then $y\geq x\geq z\geq y$, ...


2

What you have does not contain any non-trivial interval. If $0\le a<b\le 1$, there is some $n\in\Bbb Z^+$ such that $q_n\in(a,b)\subseteq[a,b]$, so clearly $[a,b]\nsubseteq G$. Whether $G$ contains any isolated points depends on exactly how $[0,1]\cap\Bbb Q$ is enumerated; I’ll sketch the construction of a $G$ that has an isolated point. Before doing so, ...


1

As you already noted, $F$ is equicontinuous and (pointwise) bounded, so Arzela-Ascoli implies its closure is compact. Let's verify $F$ is closed: Suppose $f_n$ is a sequence in $F$ which converges uniformly to some continuous function $g$. Then $f_n$ converges pointwise, hence \begin{align*} 1)&|g(x)|=\lim |f_n(x)|\leq 1;\qquad\text{and}\\ ...



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