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0

The hypothesis of continuity is necessary, and it can't be relaxed at even one point. A few counter-examples are given in the comments.


1

$f(u) = u/(1+u) = 1 - 1/(1+u)$ and $f'(u) = +1/(1+u)^2$. Hence $f$ is increasing on the domain $[0,\infty)$. Use this result and the triangle inequality on $d$ to conclude your inequality.


1

Hint: Consider the function $f(t) = \frac{t}{1 +t}$, and show that it is monotonically increasing at $[0, \infty)$.


2

Notice that the function $$f(x) = \frac{x}{1+x} $$ is increasing on $[0,+\infty[$. Indeed $$f'(x) = \frac{1}{(1+x)^2} \geq 0$$ So $f$ keep the order. And as $d(x,y) \leq d(x,z)+d(z,y)$ you have that $$f( d(x,y) ) \leq f( d(x,z)+d(z,y)) $$


1

HINT: In effect you need to prove that if $0\le a\le b$, then $$\frac{a}{1+a}\le\frac{b}{1+b}\;.$$ This is true if and only if $a(1+b)\le b(1+a)$.


1

Suppose $\int_{a}^{b}|f(t)-g(t)|dt = 0$. For any $a \le x \le b$, \begin{align} 0 \le \int_{a}^{x}|f(t)-g(t)|dt & \le \int_{a}^{x}|f(t)-g(t)|dt+\int_{x}^{b}|f(t)-g(t)|dt \\ & = \int_{a}^{b}|f-g|dt=0. \end{align} Then, the fundamental theorem of Calculus applies because $|f(t)-g(t)|$ is continuous: $$ |f(x)-g(x)| = ...


2

You are essentially asking for a proof that for continuous $h$: $[a, b]\longrightarrow\mathbb{R}$ the fact that $\int_a^b \left|h(t)\right| dt=0$ implies $h\equiv 0$. There is an elementary proof, but it's not too obvious: Suppose $h(t_0)\neq 0$ for some $t_0\in [a, b]$. Then $\left|h(t_0)\right|>0$ and hence, since $\left|h\right|$ is continuous, there ...


2

$(\Rightarrow)$ Suppose that $d(f,g)=0$. Then $$\int_a^b |f(t)-g(t)|dt=0.$$ I will call $|f(t)-g(t)|$ the (continuous) function $h(t)$. Suppose that there exists $t_0$ such that $$h(t_0)=|f(t_0)-g(t_0)|=y>0.$$ Letting $\varepsilon=y/2$, there exists $\delta>0$ (also, $\delta<b-a$) such that if $d(t,t_0)<\delta$, then ...


1

No. Take $A = (1,0)$, $B=(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}) $ , $C = (-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$ and $D=(-1,0)$ Take $d_1((x,y),(x',y')) = |x-x'|+|y-y'|$ and $d_2 = \sup( |x-x'|,|y-y'|)$ For $d_1$, the diameter is realized for B and C and for $d_2$ the diameter is realized for A and D


0

No. You could have a metric space consisting of finite set of points (but more than one) in $\mathbb R$. With the usual metric, the diameter of this set is the largest point minus the smallest point. Under the discrete metric, the diameter is 1, and could correspond to any pair of distinct points.


1

Take $x = (0, 0), y = (10, 0)$ and $z = (8, 9)$. With $p = 2$ we have a standard Euclidean distance, and $z$ is further away from $x$ than $y$ (their distances are $8 \sqrt{2}$ and $10$, respectively). With $p$ arbitrary large, however, we have the distance between $x$ and $y$ as $10$ while the distance between $z$ and $x$ as $9$.


2

Hint: Let $S = \{\sum r_k^2 : A \text{ can be covered by } (\bar B(a_k,r_k))_{k=1}^n\}$. You have shown that $S$ is non-empty. Note also that all elements of $S$ are non-negative. It now suffices to show that $S$ is closed.


2

Hint: Yes, $d$ is also a metric. As far as the triangle inequality is concerned, we have to prove \begin{equation*} \frac{\rho(x, y)}{1 + \rho(x, y)} \leqslant \frac{\rho(x, z)}{1 + \rho(x, z)} + \frac{\rho(z, y)}{1 + \rho(z, y)} \quad \forall \, x, y, z \in X. \end{equation*} Let $x, y, z \in X$ be taken at pleasure, and we define $A = \rho(x, y)$, $B = ...


0

Consider a cover of the image $F$ of $f$ (which is compact) this way: for every $x\in F$ take an open ball $B_x=B(x,\epsilon_x)$. You can take a finite subcover and let $\epsilon$ the minimum of the selected $\epsilon_x$. Now you can pick (applying the uniform continuity) $\delta>0$ such that the image of every ball with a radius lesser than $\delta$ is ...


3

HINT: Use the fact that $d$ satisfies the triangle inequality to show that if $A\subseteq B_\epsilon$, and $B\subseteq C_\delta$, then $A\subseteq C_{\delta+\epsilon}$.


3

Hint. To prove $$d_2(x,y) \leq d_1(x,y)$$ square both sides. To prove $$d_1(x,y) \leq \sqrt{k} d_2(x,y)$$ use Cauchy-Schwartz inequality as $$d_1(x,y)=\sum_{i=1}^{k}|x^i-y^i|=\vert x^1 - y^1 \vert . 1 + \dots + \vert x^k - y^k \vert . 1$$


2

No, it only holds in normable metric spaces, spaces where the metric is derived from a norm. You can take an example of the metric $d(x,y)=|x|+|y|)$. Then $d(x+z,y+z)=|x+z|+|y+z| \neq d(x,y)$ unless $z=0$.


2

No, since $+$ and scalar multiplication are not defined. By the way, they are true for normed spaces.


1

Your example is fine. Another simple one, which I think allows to visualize what happens, is as follows: let $A=\{(x,0)| x\in \mathbb{R}\}$ the real line embedded in $\mathbb{R}^2$. Then let $$U := \{(x,y)| |y|< \frac{1}{|x|}\}$$ (with the convention $1/0 := \infty$) the region between the graphs of $\pm 1/x$. The graph of $1/x$ approaches $A$ as ...


1

Hint For each point $x$ of the compact $F$ there is a ball $B_x=B(x,\epsilon_x)$ contained in $G$. Now, since $F$ is compact, you can take a finite subcover $\{B_{x_n}\}$, and consider the minimum $\epsilon_{x_n}$.


0

This is false. The property you mention for $f$ is called an open map: (i) An open map can be non-continuous: suppose $Y$ is a discrete topological space. Then any map from $X$ to $Y$ is open, but can be non-continuous. (ii) A continuous map is not necessarily open: if $Y$ is a non-open subset of $X$, with the induced topology, the canonical injection of ...


1

A function $f:X\to Y$ is continuous if and only if $f^{-1}(V)$ is open in $X$ for any open set $V\subset Y$. (This is sometimes taken to be the definition.) A function $f:X\to Y$ is open if and only if $f(V)$ is open in $Y$ for any open set $V\subset X$. (This is the definition.) These two are not the same.


0

Hard to beat the simple solution of @Keith: We'll give a proof for $X$ infinite dimensional Banach space (extra condition). First, show that there exists a sequence $x_n$ in $X$ such that $||x_n|| =1$ and $d(x_n, \langle x_1, \ldots x_{n-1}\rangle ) \ge \frac{1}{2}$. One constructs the sequence inductively. Once $x_1$, $\ldots x_{n-1}$ are obtained, take ...


2

Obviously $$ A=\left[-1,1\right]\times\left[-\frac12,\frac12\right]\times\ldots\times\left[-\frac1n,\frac1n\right]\times\ldots=\prod_{n=1}^\infty\left[-\frac1n,\frac1n\right]. $$ Given $a\in \bar{A}$, there is a sequence $(a^k)$ of elements of $A$ whose limit is $a$, i.e. $\lim_{k\to\infty}\|a-a^k\|_2=0$. Therefore, for every $k,n\ge 1$ we have: $$ |a_n|\le ...


2

Let $\{ x^{(m)} = (x_1^{(m)}, x_2^{(m)},\ldots ) \}_{m=1}^{\infty}$ be a sequence in $A$ and assume that $x^{(m)} \to x$ in $\ell^2$ as $m \to \infty$. To show that $A$ is closed, we need to show that $x \in A$. Write $x = (x_1,x_2,\ldots)$, then $x^{(m)} \to X$ in $\ell^2$ means that $$ \| x^{(m)}-x \|_{\ell^2}^2 = \sum_{n=1}^{\infty} |x^{(m)}_n - x_n |^2 ...


0

Hint: The complement of $A$ is $$ A^c=\left\{y\in l_2: |y_i|>\frac{1}{i}\text{ for some }i\right\}. $$ If you are having trouble with your method, you could show that $A^c$ is open.


5

Consider continuous functionals $$ f_n:\ell_2\to\mathbb{R}:x\mapsto x_n $$ for each $n\in\mathbb{N}$. Note that $A_n:=(f_n)^{-1}([-1/n,1/n])$ is closed as preimage of closed set under continuous map $f_n$. Since $A=\bigcap_{n=1}^\infty A_n$, then $A$ is closed as intersection of closed sets.


1

Yes, your negation of $3)$ is correct. You can think of it like this: In a not totally bounded space, there's room for the sequence to "run away". While total boundedness implies the existence of an entire Cauchy subsequence, already the seemingly weaker fact that the sequence has infinitely many pairs of arbitrarily close elements is enough to show that ...


0

Take a set $X$. Let $\lambda \in X$ , we say that a ball of radius epsilon ($\epsilon$) around the point $\lambda$ will be an open set iff $\ni$ a point $k\notin X$ such that a neighborhood of radius delta ($\delta$) around the point $k$ is not in the ball of radius $\epsilon$ around $\lambda$. In mathematical terms , $B_\epsilon(\lambda)$ is open ...


2

Let $S$ be the set of isolated points of $M$. If $S$ is infinite, then, since isolated points are open, writing $S$ as the disjoint union of two infinite sets gives what's needed. Suppose $S$ is finite. Then since $M$ is infinite, there are distinct elements $m_1$ and $m_2$ of $M\setminus S$. Choose disjoint open nhoods $N_1$ of $m_1$ and $N_2$ of $m_2$. ...


0

as fast answer I would suggest to construct by induction such set. Being that an open set is a set where every point contains an open neightbourhood completely contained insided the original set. I will try first to demonstrate that such set with an added point is a new set still open. Then by induction. The key point is the induction step


6

Every infinite-dimensional normed space has a non-closed subspace. Let $X$ be an infinite-dimensional normed space, let $a$ be a nonzero vector. Assume by induction that we have found vectors $x_1, x_2, \dots, x_{n-1}$ for which $|x_i - a| < 1/i$ and $a \not\in V_{n-1} = \Sigma_{i=1}^{n-1} \mathbf{R}x_i$. We will extend this sequence by finding an $n$th ...


3

Just take $(x,0,0,\ldots,0)$ where $x\in\Bbb R\setminus\Bbb Q$. Each point is a ball of radius $0$, clearly disjoint from every other ball, and the complement is certainly path connected.


0

First, recall the definitions: Closure: The union of a set and its limit points, or the union of a set and its boundary points, or even more broadly, $x$ is in the closure of $S$ if every open interval containing $x$ intersects $S$ Boundary point: Every neighborhood of $x$ intersects both $S$ and the complement of $S$ Interior point: at least one ...


1

Let $S=\{x_1,\ldots,x_n\}\subset \mathbb R$. If $x_j\in S$, then any open ball centered at $x_j$ contains infinitely many points, so clearly it contains a point not in $S$, and so $x_j$ isn't an interior point of $S$. Hence the interior is empty. A singleton set $\{x\}$ is closed as $x=\bigcap_{n=2}^\infty \left[x-\frac1n,x+\frac1n\right]$ is the ...


0

None of the points in your finite set $A \subset \mathbb{R}$ are interior points: no open interval around any point $x \in A$ is contained in $\{ x \}.$ This shows that the interior of $A$ is the empty set. On the other hand the set of boundary points of $A$ consists exactly of $A$: every point $x \in A$ is a boundary point since every open interval around ...


0

An open set contains an open interval and thus has infinitely many points. Thus the interior of a finite set is empty. Any point outside the set has a positive minimum distance from points in the set, so there is an open interval around it not intersecting the set. Thus the set is closed because its complement is open. A closed set with empty interior is ...


0

Yes it is. Assume you have to go from point $A$ to point $B\neq A$. If you just "go straight", you may intersect a disk $D_n$, but the intersection between your path $P$ and $D_n$ is homotopy equivalent to a small path on $\partial D_n$ (just consider the central projection with respect to the centre of $D_k$ as homeomorphism). Since the disks $D_m$ are ...


0

Having some notion of open sets lets you define notions of points getting arbitrary close to each other, or continuity. For example in space of real functions you can choose topology (in other words: choose which sets you declare open) such that sequence of functions converges to some given function pointwise. There is no natural notion of distance related ...


2

a topological vector space that is not a metric space: take $V=C(\Bbb{R})$ where the topology is given by convergence on compact sets. A basis for this topology is given by sets of the form $$U_{K,f,\varepsilon} = \{ g : \sup_K |g-f| < \varepsilon \}$$ where $f \in V$ is continuous, $\varepsilon >0$ is a positive real number, $K \subset \Bbb{R}$ is a ...


4

$\left\Vert p_{n}-p_{m}\right\Vert =\sup_{\left[ 0,1\right] }\left\vert %TCIMACRO{\dsum \limits_{i=0}^{n}}% %BeginExpansion {\displaystyle\sum\limits_{i=0}^{n}} %EndExpansion \frac{x^{i}}{i!}-% %TCIMACRO{\dsum \limits_{i=0}^{m}}% %BeginExpansion {\displaystyle\sum\limits_{i=0}^{m}} %EndExpansion \frac{x^{i}}{i!}\right\vert =\sup_{\left[ 0,1\right] ...


1

Yes, it is outer regular. First of all, since $I$ has only countably many factors, the product $\sigma$-field coincides with the Borel $\sigma$-field, that is, $({\cal B}([0,1]))^{\mathbb N}={\cal B}([0,1]^{\mathbb N})$. (see Proposition 8.1.5. on page 256 of [C]). So the product measure is a Borel measure. Secondly, every finite Borel measure on a ...


0

First question: yes. Second: That wouldn't even work, if $f$ is continuous. Think $f$ as a constant function and $g$ discontinuous.


0

If we assume for a moment that the infimum were always attained, that is, we had $$d(x,y) = \min \{ L(\sigma) : \sigma \in T_{x,y}(X)\},$$ then we could state the triangle inequality as "the shortest path from $x$ to $y$ is not longer than the shortest path from $x$ to $y$ that passes through $z$", and it would be immediate from picking paths of minimal ...


4

Edit: I've written a much simpler version of the proof of the main theorem. I will show that every completion of $\mathbb{R}$ is obtained by joining $\mathbb{R}$ to a "space at infinity", which the line "converges" to on its ends. The examples of the topologist's sine curve (together with an interval along the $y$-axis) and a spiral converging to a circle ...


0

The fact that $S$ is discrete tells you that its elements are "bounded away" from each other (if $x\neq y$ then the distance between them is more than a fixed real number, call it $\alpha$; in the discrete metric, this number is 1); this means that they cannot get arbitrarily close to each other and is the reason why a convergent sequence of elements of $S$ ...


2

He’s actually omitted a key observation: since $\sigma=\langle x_n:n\in\Bbb N\rangle$ has no convergent subsequence, it also has no constant subsequence, and therefore the set $S=\{x_n:n\in\Bbb N\}$ must be an infinite set. Thus, we might as well assume that the terms of the sequence are all distinct, i.e., that $x_m\ne x_n$ whenever $m\ne n$. Now we want ...


1

I think the "every convergent sequence with terms in the set $S=\{x1,x2,\cdots\}$ must be eventually constant" in the proof means all sequences with elements in $S$ that converge must be in this form: $$ x_{k_1}, x_{k_2}, \cdots , x_{k_n} , x_{k_n}, x_{k_n} , \cdots $$ Here you should treat $S$ as an ordinary set and we are talking about sequences whose ...


7

Any complete metric space with a dense embedding of $\mathbb{R}$. Here's a sketch of the proof: Any space that is a completion of $\mathbb{R}^1$ obviously has such an embedding (the inclusion). And if a metric space $X$ has a dense embedding of $\mathbb{R}^1$, then the induced metric on $\mathbb{R}^1$ must complete to $X$ (some sequence of points in ...



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