@psie yes

are you able to send a screenshot of the pages you're reading?

yes I think I just have to follow through with negating the definition of the interior for its complement

@psie that inequality means the identity map from $(\mathbb R^n,\|\cdot\|_a)$ to $(\mathbb R^n,\|\cdot\|_b)$ is continuous, since we can take the sup over all $\|x\|_a=1$ and get that the operator norm of the identity is bounded, hence continuous.

@SineoftheTime yes

I've worked exercise 3 already, and I'm stuck on 4.

I can also send the suggested solution to 4 in the back of the book, where they claim the unit sphere in the 1-norm is compact.

Just a second.

^ that is the solution to exercise 4, the one about showing any two norms on $\mathbb R^n$ are equivalent.

Lemma 7.2 states that the norm function is continuous.

the topology induced by the $1$-norm is the usual Euclidean topology

(but that's also to show here)

you could explicitly show that the identity from the the 2-norm to the 1-norm is continuous

you can show that if $A$ is open in the "topology $1$" then it's open in the "topology $2$" and the use HB

@BenSteffan no that's "obvious"

is it?

I guess if you look at sequences...

Maybe Thorgott has a point. There is a theorem prior to this exercise (in the section on product of metric spaces) that says that all metrics satisfying the below property determine the same family of open sets in $X=X_1\times\cdots\times X_n$.

> Property: A sequence $\{x^{(j)}=(x_k^{(j)})\}_{j=1}^\infty$ converges to $x=(x_1,\ldots,x_n)$ in $X$ iff for each $k$ the sequence of component entries $\{x_k^{(j)}\}_{j=1}^\infty$ converges to $x_k$ in $X$.

sure, you could use that

it's also easy to see geometrically: the open balls for the $1$-norm are "diamonds" (squares rotated by 45°)

TIL Emile Borel was a politician..

and it's straightforward that every ball around a point contains a diamond around that point and vice versa

ok sure

similar vibe to the heavy metal guitarist that "discovered" water on mars

you might still have to translate this into a proof :)

work seemingly worlds apart (no pun intended)

of course, I'm just here to throw around ideas

can you throw me an idea about the following: suppose you know neither about homotopy excision in any capacity nor about the Hurewicz theorem. For any given $n$, construct a CW-complex $X$ with $\pi_n(X) \cong G$ having cells only in dimensions 0, $n$, and $n + 1$ in the usual way. Can you prove that $\pi_n(X) \cong G$?

of course by "suppose you know neither" it is implied you only know about stuff that is more "basic" than this

I do fall under this category since I do not know about homotopy excision and I do not know about the Hurewicz theorem. Could I prove that thing is isomorphic to that other thing? Maybe, in a few more years.

@BenSteffan ah, then nvm. exclude me from that category

is $\{1,1/2,2/3,3/4,4/5,\dots\}$ compact? It looks to be bounded and closed

@Thorgott I have a silly question. That the topology induced by the $1$-norm is the same as the Euclidean topology...that does not mean $\{x:\|x\|_2=1\}$ and $\{x:\|x\|_1=1\}$ are the same sets, right? I guess it only means that there is a compact set in the Euclidean topology that equals $\{x:\|x\|_1=1\}$.

looks like it converges to 1

yeah I should have just gone with my instinct lol