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7

Hint: It essentialy boils down to showing that metric spaces are Hausdorff. Take $x,y \in X$ with $x \neq y$. Then $d(x,y) > 0$. Can you check that $$B\left(x,\frac{d(x,y)}{2} \right) \cap B\left(y,\frac{d(x,y)}{2}\right) = \varnothing?$$You should convince yourself with a drawing. To prove it formally, take $z$ in that intersection and get a ...


5

This is wrong. $6$ divides the sum of the digits of $15$, but $6$ does not divide $15$. $6$ divides $12$, but $6$ does not divide the sum of the digits of $12$. What is true is that $3$ divides $n$ if and only if $3$ divides the sum of the digits of $n$. $6$ divides $n$ if and only if $3$ divides the sum of the digits of $n$ and the last digit of $n$ is ...


4

The statement you are trying to prove is true, so you are unable to disprove it. Hint: The statement says that for every $x$, there exists some $y$. So all you need to do is find some particular $y$ for which the equality holds. Try to prove the statement by using very simple numbers for $y$....


4

It's overloaded notation, for sure. But I have never seen it mean anything other than the following: Let $f : X \to Y$. Then, for $A \subseteq X$, $f(A) := \{ f(x) : x \in A \}$. The reason there isn't much notice is probably because it's a very common convention. It may even be explicit at some point in some books on Real Analysis. It turns out that ...


3

Everything looks good except for when $p$ is odd, the part where you say it has at most $MN$ edges if the parts have sizes $M$ and $N$ is very good. I would proceed as follows: Clearly $M+N=p$ so we can assume $M\geq N$ and $M=\frac{p+k}{2}$ and $N=\frac{p-k}{2}$, for some positive integer $k$. The options for $k$ are $0,2,\dots,p$ if $p$ is even and $1,3,...


3

The definition of $f$ isn't changing. $f(U)$ is simply standard notation for the image of $U$ under $f$, and similarly for the inverse image $f^{-1}(U)$. Of course $f:X \to Y$ is not the same function as its direct image function, which you could denote $\mathcal{P}(f):\mathcal{P}(X) \to \mathcal{P}(Y)$ if you really wanted. If category theory interests you,...


3

Yeah, I don't see anything wrong with this line of thinking. To visit your bullet points: Allows you to solve an atomic problem well, once, and reuse it many, many times without need to revisit the details. This is more true for lemmas than theorems. If you're trying to write out a proof for a complicated theorem, the decision of which lemmas to extract is ...


2

Assume the arithmetic progression, starting with $\frac1{m_1}$, contains $k$ positive terms. Now the (uniform) term difference is $$ d = \frac{1}{m_1} - \frac{1}{m_2} \\ m_2 \geq m_1+1 \implies d \geq \frac{1}{m_1} - \frac{1}{m_1+1}=\frac{1}{m_1(m_1+1)} $$ The entire sequence, which ends in a positive value $\frac{1}{m_k}$, consists of $k-1$ steps of size $...


2

It is important in this problem to understand that the maximun number $k$ is determined by the choice of the two first integers $m_1,m_2$. Let $m_1$ and $m_2=m_1+h$ (where $h\ge 1$); the common difference of the a. p. is $d=\frac{-h}{m_1(m_1+h)}$ so we have $$u_1=\frac{1}{m_1}\\u_2=\frac{1}{m_1+h}\\u_3=\frac{1}{m_1+h}+\frac{-h}{m_1(m_1+h)}=\frac{m_1-h}{m_1(...


2

Since they are in arithmetic progression let us call $1/m_n-1/m_{n+1}=L>0$. We then have that $$\begin{align} \sum_{n=1}^{k-1}\left(\frac{1}{m_n}-\frac{1}{m_{n+1}}\right) &= \sum_{n=1}^{k-1}L \\ &= (k-1)L \\ &= \frac{1}{m_1}-\frac{1}{m_k} \end{align}$$ Which means we just have to show $$k=\frac{\frac{1}{m_1}-\frac{1}{m_k}}{L}+1<m_1+2\qquad (...


2

Some authors use different notation, i.e., for $f\colon A\to B$, the associated direct image function is denoted $f_\to\colon \mathcal P(A)\to \mathcal P(B)$. However, since this association is very common it is standrd to simply write $f$ for the extended set valued function, without warning as you say. Some care must be exercised, since for instance if $...


2

First, note that for any $x$ we have $b(x, x) \geq b(x, x) + b(x, x)$ by the second rule, and so $b(x, x) = 0$ for any $x$. Now let $x\neq y$, so that $b(x, y) = c > 0$. Then by the second and last requirements, $0 = b(x,x) \geq b(x, y) + b(y, x) = 2b(x, y) = 2c > 0$, which is impossible.


1

Suppose you have two distinct points $x, y$, then you know that $$0=b(x,x) \ge b(x, y) +b(y, x) = 2b(x,y)\ge0.$$ What does this tell you about $x, y$?


1

Let $X=\bigcap_{\alpha\in I_{1}}A_{\alpha}\times\bigcap_{\beta\in I_{2}}B_{\beta}$ and $Y=\bigcap_{(\alpha,\beta)\in I_{1}\times I_{2}}A_{\alpha}\times B_{\beta}$. Suppose $(a,b)\in X$. Then, $a\in A_{\alpha}$ for all $\alpha\in I_{1}$ and $b\in B_{\beta}$ for all $\beta\in B_{1}$. Now, let $(\alpha_{0},\beta_{0})\in I_{1}\times I_{2}$. By assumption, $a\in ...


1

There is no universal definition for "well defined." Here are three common uses of the term. $X, Y$ are sets, $\sim$ is an equivalence relation on $X$, $\tilde{X}$ is the set of subsets $S$ of $X$ having the property that $x, y \in S$ if and only if $x \sim y$. A typical element $S$ of $\tilde{X}$ is often written as $\overline{x}$, for some $x \...


1

Your interpretation of "finer" is backwards: for the uniform topology to be finer, each product-open set must be uniform-open, meaning that given any product-open neighborhood of a point we can find a uniform-open neighborhood inside it. So it would suffice to show that $B_\epsilon^u(x) \subseteq B_\epsilon^p(x)$, which you have correctly observed is true ...


1

It is an overkill, but I think it is an interesting one. $C^0(0,1)\subset L^2(0,1)$, hence we may assume that the functions we are dealing with have a Fourier-Legendre expansion in terms of shifted Legendre polynomials: they give an orthogonal complete base of $L^2(0,1)$ with respect to the usual inner product and by assuming $$ f(x)=\sum_{n\geq 0}c_n P_n(2x-...


1

Let $\{ U_{\alpha}\}_{\alpha \in I}$ be a pairwise disjoint collection of open sets in $(X, \tau)$. The index set $I$, at the outset, may or may not be countable. We will show that it is Because $X$ is separable, it has a dense countable subset $D$. Therefore, for each $\alpha \in I$, we can choose an element $u_{\alpha} \in D \cap U_{\alpha}$. Since ...


1

Your argument is essentially correct, but can be made more formal: Suppose that $\{ U_i: i \in I \}$ is a family of non-empty pairwise disjoint sets. Then define for $i \in I$, $f(i) = \min \{n: d_n \in U_i \}$, where $d_n$ is an enumeration of the dense set by $\mathbb{N}$. As $U_i$ is non-empty and open, the set of such $n$ is empty, and for ...


1

You have the right basic idea, but there’s no need to argue by contradiction. Let $D$ be a countable dense subset of $X$, and suppose that $\mathscr{U}$ is a family of pairwise disjoint non-empty open sets. For each $U\in\mathscr{U}$ there is an $x_U\in D\cap U$. If $U,V\in\mathscr{U}$, and $U\ne V$, then $U\cap V=\varnothing$, so $x_U\ne x_V$. Thus, the ...


1

It’s certainly not hard to show that $d(x,C)>0$ if $C$ is closed and $x\notin C$, but this is something that requires proof. As for the rest, just take any positive $r\le\frac12d(x,C)$ and use $B_d(x,r)$ and $$\bigcup_{y\in C}B_d(y,r)\;.$$ It also takes a little work to show that $\bigcup_{y\in\operatorname{bdry}C}B_d(y,\epsilon_y)\cup C$ is open, and ...


1

Since $r=d(x,C)$, I think that $U=B_{r/2}(x)$ and $V=\bigcup_{y\in C}B_{r/2}(y)$ should work.


1

Your approach can be carried out; you just have to figure out how to choose the points $x_n$ so that $\langle x_n:n\in\Bbb N\rangle$ is a Cauchy sequence. I’ll get you started. For $x\in X$ and $r>0$ let $C(x,r)=\{y\in X:d(x,y)\le r\}$, the closed ball of radius $r$ centred at $x$. Begin by choosing any $x_0\in U\cap D_0$; there is an $r_0>0$ such ...


1

$$(\exists\alpha_{0}\in\Lambda)\left[x\in\displaystyle\bigcup_{B\in\gamma_{\alpha_0}}B\right] \iff (\exists \alpha_0 \in \Lambda)[(\exists B \in \gamma_{\alpha_0})(x \in B)] \iff [(\exists \alpha_0 \in \Lambda)(\exists B \in \gamma_{\alpha_0})](x \in B) \iff \left(\exists B \in \bigcup_{\alpha \in \Lambda} \gamma_{\alpha} \right)(x \in B) \iff (\exists B \in ...


1

I'm not sure you need to use removal lemma to prove your claim, you can prove it using simple combinatoric argument: every edge in the graph can participate, at most, in $n-2$ triangles. so, removing any edge will reduce the number of triangles in the graph by at most $n-2$ triangles. If you remove $\epsilon n^2$ edges, it will reduce the number of ...


1

Among numbers $a-1, b-1, c-1$ there are two having the same sign, say $a-1, b-1$. In other words, $(a-1)(b-1)\ge 0.$ Multiplying both sides by a positive number $(a+1)(b+1)$ we get \begin{align*} \left(a^2-1\right)\left(b^2-1\right)&\ge 0 \\ a^2b^2-a^2-b^2+1&\ge 0 \\ a^2b^2+2a^2+2b^2+4&\ge 3\left(a^2+b^2+1\right) \\ \left(a^2+2\right)\left(b^2+2\...



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