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## Hot answers tagged metric-spaces

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 ...

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 ...

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.

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 ...

4

$\left\Vert p_{n}-p_{m}\right\Vert =\sup_{\left[ 0,1\right] }\left\vert %TCIMACRO{\dsum \limits_{i=0}^{n}}% %BeginExpansion {\displaystyle\sum\limits_{i=0}^{n}} %EndExpansion \frac{x^{i}}{i!}-% %TCIMACRO{\dsum \limits_{i=0}^{m}}% %BeginExpansion {\displaystyle\sum\limits_{i=0}^{m}} %EndExpansion \frac{x^{i}}{i!}\right\vert =\sup_{\left[ 0,1\right] ... 3 I think I can confirm your suspicion that this doesn't necessarily hold if the target space is non-Hausdorff, assuming I haven't made a mistake somewhere... Let$\mathbb{R}$be the real line in its standard topology. Let$\mathbb{R}_0$be the real line with the topology whose non-empty open sets$U$are precisely the standard open sets$U \subseteq ...

3

Hint: If $E$ is clopen (=closed and open) then so is $\mathbb R^n\setminus E$. If either of these two is empty you are done. If not, consider a line segment with points in $E$ and $\mathbb R^n\setminus E$ as end points. We may identify this with $[0,1]$ and look at $\inf$ and $\sup$ of suitable sets.

2

I'll answer your second question first. You have $f:K\times K\to\mathbb R$. The space $K\times K$ is the product of two compact sets, and is therefore compact. This is not hard to prove, and is a very simple result of the Tychonoff Theorem. Now, in order to apply the extreme value theorem, we have to show that $f$ is continuous. To do this, pick ...

2

We can use compactness of the product $K\times K$ without mentioning it. There is a sequence $(x_m,y_m)\in K\times K$ such that $$\lim_{m\to\infty}|x_m-y_m|=\sup_{x,y\in K}|x-y|$$ by the properties of the supremum. Since $K$ is compact, the sequence $(x_m)$ has a convergent subsequence, say $(x_{m_h})$. Similarly, the sequence $(y_{m_h})$ has a convergent ...

2

If I understood you well, your $\ell^0$ is the space of complex sequences that are eventually vanishing. If that is right, consider the sequence $(c_n) \in (\ell^0)^{\mathbb N}$ defined by $c_n(k) = \frac{1}{k}$ for $1 \le k \le n$ and $c_n(k) = 0$ for $k > n$. $(c_n)$ is a Cauchy sequence as for $n < m$ $$\Vert c_n - c_m \Vert^ 2=\sum_{k=n+1}^m ... 2 Obviously$$ 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 $$\| x^{(m)}-x \|_{\ell^2}^2 = \sum_{n=1}^{\infty} |x^{(m)}_n - x_n |^2 ... 2 He’s actually omitted a key observation: since \sigma=\langle x_n:n\in\Bbb N\rangle has no convergent subsequence, it also has no constant subsequence, and therefore the set S=\{x_n:n\in\Bbb N\} must be an infinite set. Thus, we might as well assume that the terms of the sequence are all distinct, i.e., that x_m\ne x_n whenever m\ne n. Now we want ... 2 Let y=x^{2}. Consider f(x,x^{2})=\frac{x^{4}}{2x^{4}}=\frac{1}{2}.So it's not continuous at (0,0). (Even it does not have a limit, you can plug y=0) 2 Let S be the set of isolated points of M. If S is infinite, then, since isolated points are open, writing S as the disjoint union of two infinite sets gives what's needed. Suppose S is finite. Then since M is infinite, there are distinct elements m_1 and m_2 of M\setminus S. Choose disjoint open nhoods N_1 of m_1 and N_2 of m_2. ... 2 In the quoted passage no assertion is made that such a set actually exists. Read "let x be Y" as "assume for the sake of argument that some object called x satisfying Y exists". In general, this can go several places. It can lead to a contradiction, proving that such an x fails to exist. Or it could show some property that all such objects must have, even ... 2 By expanding f and then re-factorizing it in a slightly different form we find$$f(x) = \sum_{i=1}^N x^2 - 2x\cdot y_i + y_i^2 = N[x^2 - 2x\cdot \overline{y} + Y] = N(x-\overline{y})^2 + N[Y - \overline{y}^2]$$where \overline{y} = \frac{\sum_{i=1}^N y_i}{N} and Y = \frac{\sum_{i=1}^N y_i^2}{N}. From this last form, remember that x^2 \geq 0, it is ... 2 a topological vector space that is not a metric space: take V=C(\Bbb{R}) where the topology is given by convergence on compact sets. A basis for this topology is given by sets of the form$$U_{K,f,\varepsilon} = \{ g : \sup_K |g-f| < \varepsilon \}$$where f \in V is continuous, \varepsilon >0 is a positive real number, K \subset \Bbb{R} is a ... 1 HINT:$$f(x) = f(\bar y) + n \,|x-\bar y|^2$$1 Note that for each n, f_n is continuous. If f_n converged uniformly on [1,\infty), then this would imply that the limit function f:=\lim_{n\to\infty} f_n is continuous. But f_n(1)=\frac12 for each n, and f_n(x)\stackrel{n\to\infty}\longrightarrow 1 for x>1, so$$ f(x)=\begin{cases}\frac12,& x=1\\ 1,&x>1,\end{cases}$$which ... 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 ... 1 I will follow your notation. If t=0, there is nothing to prove. Let's suppose t>0. Now, none of a,b,c is equal to x; thus, without loss of generality, we can suppose a,b<x. Since t=x-a=x-b, we have a=b. 1 Hint: For a fixed real number x and a number t \geq 0, how many real numbers are at a distance t from x? 1 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. 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 ... 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 ... 1 For f:\Bbb R \to \Bbb R with f(x)=x^2, we proceed as follows: Suppose f is uniformly continuous on \Bbb R. Then for every \epsilon \gt 0, \exists \delta \gt 0 such that |f(x)-f(y)| \lt \epsilon whenever |x-y| \lt \delta for any x,y \in \Bbb R. Let x=a and y=a+\frac \delta2. Then we have |x-y|=|a-(a+\frac \delta2)| \lt \delta. ... 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 ...

1

No need for Banach's theorem itself, this is an exercise that illustrates this theorem. For the first, you need to show it is a contraction, which is just a problem showing an inequality, e.g. using the mean value theorem as supinf suggests. The mean value theorem says that for fixed $x < y$ we have some $x < z < y$ such that $f(x) - f(y) = ... 1 Hint: a typical way to apply this for real functions is to use the mean value theorem:$\forall x,y \exists z \in (x,y): f(x)-f(y)=f'(z)(x-y)$. It follows that$|f(x)-f(y)| = |f'(z)| |x-y|$. Calculating the derivatives in that cases should be easy, so this helps you to proove an inequality of the type$|f(x)-f(y)| < q |x-y|$for constant$q\$. e.g. for ...

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