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6

The reason is that the intersection of infinitely many open sets need not be open: you need the set of $V_{q_k}$ to be finite in order to ensure that $V$ is actually a neighborhood of $p$, rather than merely some set containing $p$.


5

Consider the following as subspaces of $\mathbb R$ $\{0,1\}$ or in fact any finite set is compact and discrete $[0,1]$ is compact but not discrete. $\mathbb Z$ and $\left\{\frac1n:n\in\mathbb N\,\right\}$ are discrete but not compact. (But $\left\{\frac1n:n\in\mathbb N\,\right\}\cup\{0\}$ is compact and not discrete) $(0,1)$ and $\mathbb Q$ are neither ...


5

If the sequence of z differs from the sequence of x before it it differs from the sequence of y (i.e $N(x,z) \leq N (x,y)$) one has $d(x,y) \leq d(x,z)$ so the triangle inequality will be fullfiled trivially. Therefore it remains to look at the case $N(x,y) < N(x,z)$ this means, that the digits $x_n,y_n,z_n$ are the same for $n < N(x,y)$. But by ...


3

For the answer to the first question: let $\mathcal S' = \{A \in \mathcal S : A^C \in \mathcal S\}$. See that $\mathcal S'$ contains all open sets, that $\mathcal S'$ is closed under taking complements, and that $\mathcal S'$ is closed under countable unions.


3

You might consider the following image: Each of the black circles represents the minimum distance points must be away from each other. The blue circle is the unit circle. Notice that moving any single point out of this configuration necessarily either places a point into the unit circle or increases the sum of distances to the origin. Edit: There is ...


3

The map $f \mapsto \int_a^b f(x)\,dx$ is continuous with respect to the metric $d$. So if you have a sequence $f_n \to f$ converging in the metric $d$, you know that $\int_a^b f_n(x)\,dx \to \int_a^b f(x)\,dx$. In other words, convergence in the metric $d$ is exactly the right thing to guarantee that you can pass the limit under the integral sign.


3

See the first paragraph after the Remark at the top of page $9$: Since $X$ is a separable metric space, by Tychonoff’s embedding theorem, $X$ is homeomorphic to a subset of a compact metric space. Thus $X$ admits an equivalent metric $\rho$ with respect to which it is totally bounded. I would appeal instead to the Urysohn metrization theorem, but ...


3

The group of real numbers $(\Bbb R,+,0,-)$ with its usual topology is a topological group: The addition $+:\Bbb R\times\Bbb R \to \Bbb R$ (where $\Bbb R\times \Bbb R$ has the product topology) and the negation $-:\Bbb R \to \Bbb R$ are continuous maps. These properties of $\Bbb R$ are exactly what leads to formulas like $$\lim(x_n+y_n) = x+y\quad \text{ and ...


3

Hint: For any $m \in \mathbb N,$ $\Gamma$ is contained in $$\bigcup_{k=1}^m \ B(f(k/m), K/m),$$ where $B(a,r)$ is the closed ball centered at $a$ of radius $r.$


2

You were doing fine until you got to the point of showing that $\sup E\in E$. Picking the sequence $\langle r_n:n\in\Bbb Z^+\rangle$ in $E$ converging to $\sup E$ is find but then you write this: For each $r_n$ there is some $U_\alpha$ such that $B(x, r_n) \subseteq U_\alpha$ The set $U_\alpha$ must surely depend on $r_n$ and on $x$, but these ...


2

Both definitions are equivalent. Note $$B_r(a) \subset \overline{B_r(x)} \subset B_{r+1}(x)$$


2

No, $\overline{\mathbb{R}}$ with the usual distance is not a metric space. It is better understood as a topological space (with the order topology). Although it is not a metric space with the usual distance, it is metrizable. Think about a way to put a metric on it. But I will disagree with one of the comments (the one that says that ...


2

In general in analysis it is desirable to have objects of interest such as functions classified in spaces with "good properties". This allows one to use some standardised proof techniques for some problems and is very convenient. Completeness turns out to be one of the most important of such properties. Also, integration is one of the fundamental operations ...


2

Here is an example of a metric $d$ on $\mathbf R$ such that $(\mathbf R,d)$ is not complete. Set $d(x,y)=\bigl\lvert\mathrm e^{-x}-\mathrm e^{-y}\bigr\rvert$. It trivially satisfies the axioms of a metric. Now the sequence of natural numbers iss a Cauchy sequence. Indeed choose $\varepsilon> 0$ and let $N$ be an integer such that $\;\mathrm ...


2

Here is one way I like to think of it: Suppose you're trying to cover an infinite compact set, and you really want to give it an infinite cover that doesn't have a finite subcover. So you get a collection of infinite sets that looks like it nearly covers everything - maybe you've left behind a countable subset of an uncountable set or something. You must ...


2

Your examples: The discrete metric space on a finite set is compact. Closed bounded sets in $\mathbb{R}^n$ are compact. The discrete metric space on an infinite set is not compact. Many examples in $\mathbb{R}^n$ are available here, but open balls are probably the most easily visualized. If your goal is to study metric spaces rather than topological ...


1

There are different ways to do that. You could use euclidean distance (or any other metric applicable to vector spaces) if the alphabet is in a metric space. As a special case if you use $L^1$ distance and your alphabet is ${0, 1}$ with usual metric you will get the hamming distance. Since you're talking permutation distance, then yes it's a metric too ...


1

The first statement is true both ways. Specifically, suppose $(X, ||\cdot||)$ is a normed linear space. Then the norm $||\cdot ||$ is induced by an inner product iff the parallelogram law holds in $(X,||\cdot||)$. For the second statement, this is not true. Call the condition $d(x,y)=d(x+a,y+a)$ translation invariance, and the condition $d(x,y)=d(ax,ay)$ ...


1

The target function is continuous and we can restrict the domain to points in the compact annulus $1\le r\le 3$. Hence the minimum of the function is attained. Therefore it suffices to show for any configuration that is not a regular decagon centered at $O$ and with side length $1/\sqrt 2$, there exists a better configuration. This way you can readily show ...


1

Let $$m_0 := \{t \in \mathbb{R}^\mathbb{N} : \{t_1, t_2, \ldots \} \text{ is finite} \}$$ be the sequences of real numbers that only take on finitely many values. Equip it with the sup-norm, i.e. $$\| t \|_\infty := \sup_{i \in \mathbb{N}} |t_i|.$$ This is an incomplete normed vector space (so it's also a metric space). To see this, consider the sequences ...


1

For a non-metrizable but locally compact Hausdorff counterexample, consider $\omega_1$ with its order topology. I will show that $\mathcal{S}$ does not even contain all the open sets. Let $\mathcal{B}$ be the collection of all sets $B \subset \omega_1$ such that either $B$ is countable or $B$ contains a club set. (We consider "countable" to include ...


1

You can proof that they are equivalent definitions.


1

Indeed the algebra of limits holds (for sums and scalar multiples) in any normed space. The proof is identical. The remaining algebra of limits hold in any normed algebra. Note here that the norm on an algebra imust satisfy $\lVert xy\rVert\leq\lVert x\rVert\lVert y\rVert$ for all $x,y$ in the algebra. Also note that non-zero must be replaced by invertible ...



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