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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 ...


4

The open ball $\{y\in X : d(x,y) < \frac{1}{2}\} = \{x\}$ Hence every single point set is open. Can you proceed?


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.


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, ...


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 $f(x)=\phi(d(x,A))$.


2

That $\bigcap C\subseteq\operatorname{iso}(X)$ easily follows from the following: Claim: If $x\in X$ is not isolated, then $X\setminus\{x\}$ is dense in $X$. Proof: Suppose that $x\in X$ is not isolated. Let $V\subseteq X$ be a neighborhood of $X$. By definition, $V$ contains a point other than $x$, so $V\cap (X\setminus\{x\})\neq\varnothing$. This means ...


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 ...


2

Concerning the usage of "if": In definitions people usually write X is called Y if Z holds, even though they mean "if and only if". Since it is a definition, there is no other object with that name and other probably weaker definitions. Concerning "if", "iff", and $\implies$: "If condition A holds, then statement B is valid" is usually written as $A ...


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 ...


2

A simple approach would be to calculate the slope of the interpolating line between every successive pair of points, then take the standard deviation of the slopes.


1

Regard your data points as simple support points through which an elastic thin beam has to pass. The beam deforms to a curve that minimizes the energy required for bending it. You might be interested in that amount of energy. It can be used to measure the beam curve's deviation from a straight line. The simplest theory for thin elastic beams is Bernoulli's ...


1

$\int_0^1\vert f_n(x)-1\vert dx=\int_0^{1/n}\vert f_n(x)-1\vert dx+\int_{1/n}^x\vert f_n(x)-1\vert dx$ 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 $0\leq x\leq\frac{1}{n}$, $f_n(x)=nx\leq 1$.


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 ...


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, $B$ ...


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}\\ ...


1

By definition an element $x$ is isolated if $\left\{ x\right\} $ is an open set. Let $x$ be isolated. If $D$ is dense then $x\in\overline{D}$ so any open set containing $x$ has a nonempy intersection with $D$. That results in $\left\{ x\right\} \cap D\neq\varnothing$ or equivalently $x\in D$. Conversely if $x$ is an element of each dense set then the set ...


1

That is correct, but it's not really 'another way'. What you've done is taken the first proof and expressed it in set-theoretic language. This is often a useful thing to do, though the first proof has the advantage that everything is explained in words rather than with symbols.


1

What happens is the following: first we state that $\sup_{\| x\|\leq S}\|Ax \|=\sup_{\| x\|= S}\|Ax \|\leq r \tag 1 $ then we see that for $x\in X$ the following holds, since $A$ is linear $$ \|Ax\|= A\left (S^{-1}\|x\|\frac{Sx}{\|x\|}\right )=S^{-1}\|x\|\|A(\frac{Sx}{\|x\|})\| $$ then we see that $\forall x\in X: \|\frac{Sx}{\|x\|}\|=S$ which means we ...


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 $A$.



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