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We can also see that $A$ is uncountable. Suppose that $A$ is countable and $A=\left \{ x_n:n\epsilon \mathbb{N} \right \}$, then we write $\left [ 0,1 \right ]=A\cup \left (\left [ 0,1 \right ]\setminus A \right )=\left (\bigcup_{n=1}^{\infty}\left \{ x_n \right \} \right )\cup \left (\bigcup_{j=1}^...

Hello everyone. For exercise (2): (i) We have that $\lambda (A_\varepsilon )=\lambda \left (\bigcup_{n=1}^{\infty}\left ( q_n-\frac{\varepsilon }{2^{n+1}},q_n+\frac{\varepsilon }{2^{n+1}} \right ) \right )\leq \sum_{n=1}^{\infty}\lambda \left ( q_n-\frac{\varepsilon }{2^{n+1}},q_n+\frac{\varepsilon ...

First of all, $H$ contains the identity permutation, is closed under multiplication and contains every element's inverse, so it is in fact a group. Obviously, we are not going to prove the result using the definition. Observe that $H$'s elements are exactly all the possible permutations of $S_4$ wri...

Another solution: We define $X_1=\left \{ 1_G \right \}, X_2=\left \{ x\in G: x\neq 1_G, x=x^{-1} \right \}$ and $X_3=G\setminus (X_1\cup X_2)$ is the set of the rest of the group's elements. These sets make up a partition of $G$ due to the uniqueness of element order and thus $G$ can be written as ...

This exercise is trivial using Cauchy's theorem, which is kind of an overkill in this situation. Here is a more elementary proof: The idea is to pair every element of the group with its inverse. Setting aside the identity element, $2n-1$ elements remain, which is an odd number of elements. So we sep...

It's due time. Solutions are given for both exercises: For the first exercise, denote this particular $H$ as $U_n(R)$. Facing an $n\times n$ matrix as a $3\times 3$ block matrix we write: $U_n(R)=\left \{ \begin{pmatrix} 1 &A &b \\ 0_{n-2} &I_{n-2} &C \\ 0 &0_{n-2} &1 \end{pm...

It is fairly easy to check (and maybe come up with) that the center of $H$ is $Z(H)=\left \{ \begin{pmatrix} 1 &0 &b \\ 0 &1 &0 \\ 0 &0 &1 \end{pmatrix}:b\in \mathbb{Z} \right \}$. Define $f:H\rightarrow \mathbb{Z}\times \mathbb{Z}$, $\begin{pmatrix} 1 &a &b \\ 0 &...

Let $(X,d)$ be a metric space. The idea is to find a metric space $(Y,d')$, such that $d'$ is a bounded metric, equivalent to $d$ and then find a homeomorphism $f:X\rightarrow Y$. Pick $Y=X$ and define $d'=\frac{d}{1+d}$, which is bounded since $d'\leqslant 1$. Step 1 : $d'$ is a metric. Obviously $...

By the Sylow theorems, Sylow $p$-subgroups exist, since $p$ divides $|G|$. We observe that for every $P \in Syl_p(G)$, we have $f(P) \in Syl_p(G)$, since automorphisms preserve orders. *If there is only one Sylow $p$-subgroup, call it $P$, we have $f(P)=P$. *In the general case, let $P_1,...,P_m$ be...