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0

No, in general this is not true. Take $A=\{0\}\subset R^n$ which is convex and compact (and let $n>0$). Then $int(A)=\emptyset$ and so is $\overline{int(A)}$, hence $\overline{int(A)}\neq A$.


1

To see that $S_1 \cap S_2 = \emptyset$ you only need to remark that an isolated point of $S$ (i.e. a point in $S_2$) is certainly not a limit point of $S$, i.e. a point of $S_1$. Your other argument goes a long way in proving the equality $\overline{S} = S_1 \cup S_2$, though. First, $S_1 \subset \overline{S}$, trivially, and $S_2 \subset S$, so certainly ...


0

In a metric space $X$, the closure of $S$ is the set of all points $x \in X$ such that $x_n \to x$ for some sequence $\{x_n\}_n$ of points from $S$. Be careful, the constant sequence is allowed, and that's why you need isolated points (belonging to $S$, of course).


0

It depends on your norm on $W^{m,p}(\Omega)$. You could choose \begin{equation*} \| u \|_{m,p}^2 = \sum_\alpha \|D^\alpha u \|_p^2, \end{equation*} or \begin{equation*} \| u \|_{m,p}^p = \sum_\alpha \|D^\alpha u \|_p^p. \end{equation*} Then, you similarly choose \begin{equation*} \| u \|_{p}^2 = \sum_{i=1}^N \|u_i \|_p^2, \end{equation*} or ...


2

Look at the $\epsilon$-Balls generated by thus norm, i.e. at $$ B^n_\epsilon(x) = \{y \,:\, \|x - y\|_p < \epsilon \} \text{.} $$ These "balls" are $n$-dimensional rectangles, i.e. $$ B^n_\epsilon(x) = x + (-\epsilon,\epsilon)^n \text{.} $$ Now look at the open sets in the product topology on $\mathbb{R}^n$. This topology is generated by the base $$ ...


1

The topology of $\mathbb{R}$ has the set of open intervals in $\mathbb{R}$ as a basis. Since the product topology on $\mathbb{R}^n$ has $\{U_1\times U_2\times...\times U_n|U_1,...,U_n\text{are open in $\mathbb{R}$}\}$ as a basis, one can show that $\{]a_1,b_1[\times ]a_2,b_2[\times...\times ]a_n,b_n[\,|a_1,..,a_n,b_1,..,b_n\in\mathbb{R}, ...


1

As pointed out in the comments, $d$ will become a metric if we rather consider equivalence classes of functions. It has been shown in this website that this metric is equivalent to convergence in measure, i.e. $d(f_n,f)\to 0$ if and only if $f_n\to f$ in measure. Here we can rediscover the fact that if a sequence converges in measure, a subsequence ...


15

Metric spaces were introduced by Frechet in his PhD dissertation on functional analysis, in 1906. Functional analysis (and rigorous modern analysis) was still quite new at the time. Also, an abstract, axiomatic, approach to mathematics was also not yet as routine as it is for us today. The mathematicians of that time were studying various spaces (mainly ...


7

You may have been introduced in your analysis class to the notion of convergence, but only about the convergence of real sequences. If we want to extend this idea for say vectors, complex numbers, functions, or even sequences of sequences, and to generalize it, we can define for every mathematical object the meaning of convergence. But as you may guess, ...


6

Note that $$[x]=\sqrt{(x_1+{\textstyle\frac{1}{2}}x_2)^2+({\textstyle\frac{\sqrt3}{2}}x_2)^2}\ .$$ If you are permitted to use in your proof the triangle inequality for the standard Euclidean norm, you have $$\sqrt{(u_1+v_1)^2+(u_2+v_2)^2}\le\sqrt{u_1^2+u_2^2}+\sqrt{v_1^2+v_2^2}$$ for any real $u_1,u_2,v_1,v_2$. If you now substitute ...


1

Let $A = \bigcup_{n=1}^{\infty} (0, 5 - 1/n)$. Then $\sup (A) = 5$ by using the definitions, but $5 \notin A$.


1

$\sup$ means supremum, which is the least upper bound of a set. So for your example, $$\sup\{ |F_X(x) - F_y(x) | : x \in R\}$$ would return the least upper bound of $|F_X(x) - F_y(x) |$ for all $x\in{R}$.


2

A few things that come to mind: The topology of uniform convergence is too fine on e.g. $C(X)$ where $X$ is an open subset of some $\mathbb{R}^n$. In many situations, you don't have uniform convergence, but locally uniform convergence, and the locally uniform convergence is sufficient for many theorems (the limit of a locally uniformly convergent sequence ...


2

Just take $A = X$, but NOT with metric $d$. Take $A = X$ with a metric $p$ that gives the product topology, like $p = \max d_i$ or $p = \sum d_i$. And take $f = \mathrm{id}: (X, p) \rightarrow (X, d)$. Now, $\pi_i \circ \mathrm{id}: (X,p) \rightarrow X_i$ is certainly continuous, since $(X,p)$ has the product topology. So, you hypothesis imply that $f = ...


5

This is false. Take universal cover $M$ of the open 2 dimensional Euclidean unit disk $D^2$ with center removed, equipped with pull back flat metric. I let $d$ denote the distance function on $M$ induced by the pull-back Riemannian metric. I claim that the diameter of $M$ is $\le 2$ (actually, it is exactly 2 but we do not need this). To see this it is ...


0

Since $X$ and $Y$ are metric spaces, continuity is equivalent to sequential continuity. Let $(x_n)_n$ be a sequence in $X$ which converges to some $x\in X$. We have to show that $f(x_n)$ converges to $f(x)$ in $Y$. It suffices to show that every subsequence $(x_{n_k})_k$ has a further subsequence $(x_{n_{k_i}})_i$ such that $f(x_{n_{k_i}})$ converges to ...


1

Assume $d(z,w)=0$ and $z\not=w$. Then $$0=d(z,w)=|z|+|w|$$ If the sum of two non-negative numbers is $0$, then they are both $0$. That is $|z|=|w|=0$, i.e. $z=w=0$. Contradiction.


1

This is not true. Just pick some $K$, like a closed ball, and let $A$ and $A_n$ be whatever compact subsets of $K$ you want. Then trivially $A_n\cup K\to A\cup K$.


2

Let $y\in X-\lbrace x\rbrace$. Then $y\neq x$ and thus there is an open $U_{y}$ such that $y\in U_{y}$ and $x\notin U_{y}$. Thus $U_y\subset X-\lbrace x\rbrace $. Thus it's open because it contains an open neighbour for every element.


1

Set $$K(F,G) := \inf\{\varepsilon>0; G(x-\varepsilon) \leq F(x) \leq G(x+\varepsilon) \, \text{for all} \, x \in \mathbb{R}\}.$$ It is not difficult to see that $K$ is a metric. Let us consider the following easy example in order to see the differences between the Lévy metric $L$ and the metric $K$: Define random variables $X:= 1_{[1/2,1]}$ and $X_n ...


3

Set $F_1=X$. Then $F_1$ is compact. Set $F_2=f(F_1)=f(f(X))$. Then $F_2$ is compact because continuous image of a compact set is compact and also $F_2\subset F_1=X$. $F_3=f(F_2)$ and $F_3= f(F_2)\subset f(F_1)=F_2$ By induction prove that there exists a decreasing sequence ($F_n$) of compact sets. Then $\bigcap_{n} F_n=A\neq \emptyset$. Then ...


1

Hint: If $\displaystyle f : x \mapsto \frac{1}{x-a}$, what is $d_{\infty}(f,0)$?


1

Suppose r < y < s. The basic open set (p,q) $\times$ (r,s) contains the line (p,q) $\times \{y\}$. Is it possible that this line is contained in that subset of $\mathbb{R}^2$?


1

This is true for length/path metric spaces and false in general.


1

If $\{x_k\}$ is Cauchy in the metric $\rho$, then $\{\ln(1 + x_k)\}$ is Cauchy in the standard metric. Thus there exists a real number $y$ with the property that $\ln(1 + x_k) \to y$. Show there exists $x \in [0,\infty)$ with $\ln (1+x) = y$ and $\rho(x_k,x) \to 0$.


2

The intended metric is certainly the usual metric on the real numbers: $d(x,y) = |x-y|$. You are looking to find a function $f$ satisfying $|f(x) - f(x_0)| = |x - x_0|$ which reduces in this case to $|f(x) - x_0| = |x - x_0|$. You can solve this for $f(x)$.


0

In this answer $d\left(f,g\right)$ is an alternative notation for $\left|f-g\right|_{u}$. Define $\left\Vert f\right\Vert :=\sup\left\{ \left|f\left(s\right)\right|\mid s\in S\right\} $ and note that $d\left(f,g\right)=\left\Vert f-g\right\Vert $. When it comes to the triangular inequality then it is enough to prove that $\left\Vert f+g\right\Vert ...


1

Another counterxample is $\dfrac 12 (x+\sqrt{x^2+1})$.


2

Wikipedia's article on the Banach fixed-point theorem gives the counterexample $$ T:[1,\infty)\to [1,\infty)\qquad T(x) = x+\frac1x $$ For $X=\mathbb R$ we could take $f(x)=\sqrt{x^2+1}$.


1

Hint: $[0,\infty)$ is a closed subspace of $\mathbb{R}$, which is complete, and for $(0,\infty)$, take the sequence $\{1/n\}_{n\geq 1}$, which you can show to be Cauchy. Theorem: Let $(M,d)$ be a complete metric space with metric $d$, and let $(S,d)$ be a subspace of $(M,d)$. Then if $S$ is closed in $M$, then $(S,d)$ is complete." Proof: Suppose $S$ ...


1

It is false in metric spaces in general. For example, suppose only two points exist in the space and the distance between them is $1$. Then the open balls of radius $3/4$ about the two points are disjoint even though $3/4+3/4>1$. So you need some additional assumptions about the space involved beyond what things like the triangle inequality can give ...


2

The set $S$ should depend on a particular $ϵ$, so $$S_ϵ=\{b\in F\mid \exists\{z_o,...,z_n\}\subseteq F,\ z_0=a,\ z_n=b,\ d(z_n,z_{m+1})<ϵ\}$$ You can show that $S_ϵ$ is open in $F$: If $b\in S_ϵ$, and $(z_0,...,z_n)$ is the sequence from $a$ to $b$ in $F$, then the ball $B_{ϵ-d(z_n,z_{n-1})}(b)$ is contained in $S_ϵ$. $S_ϵ$ is also closed in $F$: If ...


0

Your approach seems correct. Your set contains the point $a$, so it's nonempty. For the openness pick any point and show that an open ball of radius $\varepsilon$ is contained in the set too. For the closedness pick a point in the closure and show there's a point from the set that is $\varepsilon$ close to it. So your set is a nonempty clopen subset of $F$ ...


1

Here is an long alternative to Stefan's succinct proof: If $a,b \in F$ and $\epsilon>0$, let an $\epsilon$-path be a finite sequence of points satisfying the condition in the question. Suppose $F$ is connected. Let $a \in F$ and $\epsilon>0$. Let $U_a(0) = B(a,\epsilon) \cap F$, and $U_a(n+1) = \cup_{x \in U_a(n)} B(x,\epsilon) \cap F$. Note that ...


0

$$|x-y| \le |x-z| + |z-y| \le |x-z| + 2*\sqrt{|x-z||z-y|} +|z-y| = \left(\sqrt{|x-z|} + \sqrt{|z-y|} \right)^2$$ Now take the root again.


1

Notice $( d(x,y) )^2 = |x-y| \leq |x-z| + |z - y| = (d(x,z))^2 + (d(z,y))^2$ Claim: If $a,b \geq 0$, then $\sqrt{a + b } \leq \sqrt{a} + \sqrt{b} $. This follows from the fact that $2 \sqrt{ab} \geq 0 $. Adding $a+b$ in both sides give: $$ a + 2\sqrt{ab} + b \geq a +b \iff (\sqrt{a} + \sqrt{b})^2 \geq a + b \iff \sqrt{a} + \sqrt{b} \geq \sqrt{a+b} $$ ...


4

Hint: We already have the usual triangle inequality $\lvert x-y\rvert\leq\lvert x-z\rvert+\lvert z-y\rvert$. This tells you that $$ \sqrt{\lvert x-y\rvert}\leq\sqrt{\lvert x-z\rvert+\lvert z-y\rvert}. $$ From here, it sufficies to prove that $\sqrt{a+b}\leq\sqrt{a}+\sqrt{b}$ for non-negative $a,b$.


0

The beautiful book: Jänich, K., Topology. Springer, 1984 also has a great picture illustrating the proof of Urysohn's lemma.


0

For your proof: what guarantees that taking the finite subcovering of W you have remaining elements of $U$? Second here is what I think is a proof: Take a covering of $C' \cap C$ call it $U$. Take $V=U \cup \{X-C\cap C'\}$ which is a open covering of $C$ (that is a consequence of Hausdorff because the fact that $C$ and $C'$ are compact implies closed ...


1

Here is an example of two compact subsets in a non-Hausdorff space whose intersections fails to be compact: Let $ℚ$ be the equipped with the topology where the only open sets in $ℚ$ are either $ℚ$ itself, or subsets of $ℤ$. Clearly, $ℚ$ and any proper superset of $ℤ$ is compact, because any open cover of such a set must contain $ℚ$ itself. But $ℤ/3 ∩ ...


2

In Hausdorff spaces compact sets are closed and in compact spaces closed sets are compact. $C$ and $C'$ are compact subsets of Hausdorff space $X$ hence are closed. Then $C$ equipped with the subspace topology is a compact space. $C'\cap C$ is closed in this compact space, hence is compact.


3

Since $C$, $C'$ are compact and $X$ is a hausdorff space, they are closed. Therefore $C\cap C'$ is closed. Now, closed subsets of compact sets are compact.


1

Hint: Consider two disjoint half-planes connected by a "bridge" that goes each time farther towards infinity. (I'm trying to be not very precise since this is homework, but if you still have questions I can elaborate)


1

Consider the sequence of annuli $A=\{1\le |z|\le 1+1/n\}$ in the complex plane. Their limit in the GH-sense is the unit circle. Since the annuli and the circle have the same cardinality, you can assume (although it is very unnatural, GH-convergence is set up to avoid this) that the corresponding metrics are defined on the same set. If you want Riemannian ...


0

The most natural thing to do is just treat the same definition for where you allow the empty set. You get that $H(X,\emptyset)=H(\emptyset, X)=\infty $ for all non-empty $X$ and that $H(\emptyset,\emptyset)=0$. It's not a special case or anything, just the definition. The thing is we don't really care about the distance from the empty set. It's not that it ...


0

I think what you want to show is $$\forall r,x \in \mathbb{R} \quad \textrm{with}\quad r>0 \quad \exists p,q \in \mathbb{Q} \quad \textrm{with}\quad p>0 \quad \textrm{such that} \quad B_p(q) \subset B_r(x).$$ Ineed, choose $q \in B_{r/2}(x) \cap \mathbb{Q}$ and $p \in (0,r/2) \cap \mathbb{Q}$ which exist by density of $\mathbb{Q}$ in $\mathbb{R}$.


1

To show that the closure of $c_{00}$ is $c_0$ you need to do precisely what Daniel suggests. a) $c_0$ is closed: Indeed let $(x^n) \subset c_0$ be a sequence (of sequences, each indexed in the following way: $(x_i^n)_{i=1}^{\infty}$) converging to some $x \in l^{\infty}$. We must show that in fact $x \in c_0$. So we must show that $\lim_{i \rightarrow ...


0

Consider a point in $c_0$ that is $x = (x_1, x_2, \dots )$ s.t. $x_n \rightarrow 0$. It is the limit of $Y_1 = (x_1, 0, 0, \dots)$ $Y_2 = (x_1, x_2, 0, 0, \dots)$ $\dots$ Check : $Y_N \rightarrow x$ in $l^\infty$ norm. So $c_0 \subset \overline{c_{00}}$ under $l^{\infty}$ norm. Is one side of the proof. For the other part consider an element $X = ...


1

With the usual metric on $\mathbb{R}$, Let $ X=\mathbb{R}$ and $A= \mathbb{Q}$.


0

Hint: In $d_{T}$ what is the largest term in the sum?



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