# Tag Info

24

why do I have to restrict the measurement standard to real number? Why not any ordered field? You can use any ordered field, and the axioms will still make sense. The thing is, though, that most of our geometric intuition is built on the Archimedian property. But if we include this in our field, then it becomes a subfield of $\Bbb R$. In this case, we ...

5

You can define inner products for vector spaces and normed vector spaces over any ordered field. You can define metrics over any ordered field: they are called generalised metrics.

5

A very important difference between metric spaces and topological spaces is that metric spaces have large-scale structure while topological spaces have only local structure. (This is only true for the right interpretation of large-scale structure, in particular, one which doesn't include compactness.) This is the motivation for the very interesting field of ...

4

I'll give a somewhat unorthodox answer, deliberately interpreting the meaning of metric space in the broadest reasonable way possible. In Flagg's "quantales and continuity spaces" the concept of value quantale is introduced. A value quantale as an abstraction of the essential structure of the non-negative reals required to define a metric space. Then, given ...

3

Apply the definition of a Cauchy sequence to $\frac{\epsilon}{2}$. So there exists $N \in \mathbb{N}$, such that for all $n,m \ge N$, $d(x_n, x_m) < {\epsilon \over 2}$. What ball of radius $\epsilon$ does the tail lie in? The last part follows because of the following fact:if a subsequence of a Cauchy sequence converges to $p$, the whole sequence ...

3

Suppose that $X$ is a metric space, and that $\langle x_n:n\in\Bbb N\rangle$ is a Cauchy sequence in $X$ that does not converge. Without loss of generality we may assume that $x_m\ne x_n$ whenever $m,n\in\Bbb N$ and $m\ne n$. Let $D=\{x_n:n\in\Bbb N\}$; then $D$ is a closed discrete set in $X$, so the map $f:D\to\Bbb R:x_n\mapsto n$ is continuous. By the ...

2

Cauchy sequences, complete space, sequential compactness as compactness, (no) real need of separation axioms less than $T_6$...

2

Without attempting to be exhaustive, in alphabetical order: balls boundedness and total boundedness Cauchy sequences/nets (make sense in topological vector spaces, too) completeness contractions Hausdorff distances isometry Lipschitz continuity uniform structures Also, the following properties are often useful: Every metric space is perfectly normal. ...

2

The statement you say you have been able to prove is correct. In fact, it's very close to a characterization of Banach spaces. A normed linear space $X$ is Banach if and only if every absolutely summable series in convergent, i.e. if every series $\sum_{n=1}^\infty a_n$, $a_n\in X$ with $$\sum_{n=1}^\infty \Vert a_n \Vert_X < \infty$$ converges in $X$. ...

2

$\Bbb Q$ with the usual distance, and the sequence $\left(1+\frac1n\right)^n$.

2

Here is an example that comes from number theory rather than from analysis. Euler believed that if you fix a prime number $p$ then infinite series of the form $$a_0 + a_1 p + a_2 p^2 + a_3 p^3 + \cdots$$ make some kind of mathematical sense, where the coefficients are chosen in $\{0,1,…,p-1\}$. More generally one might wonder how to make sense of a similar ...

2

As Bungo said in the comments, you can also consider the $\ell^p(\mathbb{N})$ spaces. For $1 \leq p < \infty$ they're the sets of summable sequences to the p-th power:$$\ell^p = \bigg\{x \colon \sum_{n = 1}^{\infty} |x_n|^p < \infty\bigg\}$$ With the norm $$\|x\|_p = \bigg(\sum_{n = 1}^{\infty} |x_n|^p\bigg)^{1/p}$$ And for $p = \infty$ ...

2

So you've got that $(X,d')$ is separable. So there exists a countable dense subset $A\subset X$ (this in fact means (according to the definition of a topology generated by metrics) that each $\varepsilon$-ball with respect to $d'$ contains a point from $A$). Take this subset and show that it's going to be dense if you consider balls with respect to $d$ ...

2

Let $X = \mathbb{R}^n$ and $d$ the usual euclidean metric. Define $d'(x,y) = \frac{d(x,y)}{d(x,y) + 1}$ for $x,y \in \mathbb{R}^n$. As the map $\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}, x \mapsto \frac{x}{x + 1}$ is monotone and concave, it can be shown that $d'$ is a metric on $X$. Notice that $B_{d'}(x_0,r) = X$ for all $r \geq 1$. We thus ...

1

Finally I found an useful result for my work. If $f$ is a real function over a polish space then $f$ is a Baire 1 function if and only if the restriction to a closed subset has a continuity point. That is the Baire Characterization Theorem. If a function is upper semi-continuous also it is Baire 1. The same for a difference of two upper semi-continuous real ...

1

Not any ordered field possesses the square root operation defined wherever $x ≥ 0$; rational numbers form an obvious counter-example known from ancient times. Yes, it is essentially the same problem as famous one faced by Pythagoreans, but in modern formulation: how to define number field to make things going well with geometry? There can be numerous ways ...

1

There are no more than one fixed point. Indeed, suppose that there are two: $x$ and $y$. Then $$d(x,y)=d(T^n(x),T^n(y))<c\cdot d(x,y)$$ and that implies $x=y$. Now take any point $x_0$, and define $x_{k+1}=T^n(x_k)$. For any pair of natural numbers $p>q$ we have $$d(x_p,x_q)<c^qd(x_{p-q},x_0)\leq ... 1 There is a typo in your definition of the metric (check the sum index), but assuming you meant to define$$ d_C(f,g) = \sum_{k=1}^\infty \max_{x \in [0,k]} 2^{-k} \vert f(x) - g(x) \vert $$then$$ \vert e_x(f) - e_x(g) \vert = \vert f(x) - g(x) \vert \le 2^x \max_{y \in [0,x]} 2^{-x} \vert f(y) - g(y) \vert \le 2^x d_C(f,g). $$1 (0,1) with the Euclidean metric (i.e. the usual metric). The sequence <\frac{1}{n} : n \in \mathbb{N}> is a Cauchy sequence in (0,1), but has no limit in (0,1). 1 Try$$\frac{\lfloor N\sqrt{2}\rfloor}{N}$$in \mathbb{Q} as N\to \infty. This converges to \sqrt{2}, which is not in \mathbb Q. 1 Let \epsilon_n = 1/n. Let A_n be the closed interval [-n,n]. Let S be any compact set. For each \epsilon_n, the Weierstrass Approximation Theorem states that there exists a polynomial P_n(x) such that for any x \in A_n,$$|f(x) - P_n(x)| < \epsilon_n The fact that $S$ is compact allows us to apply the Heine-Borel Theorem, which ...

1

Your proof is fine. (It's just that by personal taste I tend to try and use only a single, common definition for successor and limit ordinals)

1

For part a) fix the relevant $\epsilon$ and use the definition of Cauchy. You should easily see that the some tail is contained in some epsilon ball. For part b) consider $\{1/n:n\in \mathbb{N}\}$.

1

If $r = d(x,q)$ then $N_r(x)$ contains points arbitrarily close to $q$. In particular, $N_r(x)$ contains points not in $E$. Try taking a smaller $r$. Edit: Here's a specific example with more details: Take your metric space to be $\mathbb{R}$ with the standard metric. Suppose $p = 2$ and $q = 0$. Then, the set $E$ consists of the points $x$ with $|x - 0| ... 1 The property of a point$x$that for all$\varepsilon>0$, the intersection$B(x,\varepsilon)\cap S$is non-empty is equivalent to every neighborhood$U$of$x$intersects$S$since$B(x,\varepsilon)$is a neighborhood, and every neighborhood$U$contains a ball$B(x,\varepsilon)$for some$\varepsilon>0$. These characterizations define$x$... 1 Suppose that$\|p-p_0\|\le\delta_0$, and$\delta>0$; we want to find a point$q$such$\|q-p_0\|<\delta_0$and$0<\|q-p\|<\delta$. Assume for now that$p\ne p_0$; we’ll take care of the special case$p=p_0$at the end. A good place to look for$q$is on the line segment$\overline{p_0p}$, which consists precisely of the points$tp_0+(1-t)p\$ such ...

Only top voted, non community-wiki answers of a minimum length are eligible