1

skip

2

Use a strong form of Zorn

1. Zorn

2. Zorn applied to \(\mathcal{P}(S)\)

3

Show \(\mathbb{R}^\mathbb{N}\) does not have a countable basis.

Following hint for each \(x\in\mathbb{R}\) we look at the vector \[ v_x = (1, x, x^2, x^3,\dots)\] and we claim that the set \(\left\{V_x\right\}_{x\in\mathbb{R}}\) is linearly independent.

Suppose not, then there exists \(c_1,\dots,c_n\) coefficients such that \[c_1 v_{x_1} + \dots + c_n v_{x_n} = 0\] but this implies that \(v_{x_1}, \dots, v_{x_n}\) are linearly dependent, in particular this Vandermonte matrix \[ \begin{pmatrix} 1 & x_1 & x_1^2 & \dots & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \dots & x_2^{n-1} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_n & x_n^2 & \dots & x_n^{n-1} \\ \end{pmatrix} \] has determinant \(0\) despite \(x_i \ne x_j\) for all \(i\ne j\).

4

\(\ell^\infty\) is definitely equivalent to neither \(\ell^1\) or \(\ell^2\).

Constant bound between sum and sum of square seems impossible as number of summands, despite being finite, is still unbounded.

In fact from this stackexchange post sum of squares is below sum squared which is below n sum of squares.

5

5.1

obvious

5.2

positive check, definite check, scalability follows from that of \(\left\lvert-\right\rvert\) and properties of integration, triangle also follows from abs on \(\mathbb{R}\) and properties of integration.

6

6.1

trivial

6.2

1. too hard

2. by geometric intuition if \(U = V\) then \(\lVert P \rVert = 0\) otherwise \(1\)

3. too hard