If $B$ is an integrally closed domain and $B to A$ is an integral extension, then $A$ is the integral closure of $B$ in $K(A)$ or not?

So i found a riddle on reddit which is a consequence of the axiom of choice.

There is a house with 100 rooms, and each room contains countably many boxes indexed with the natural numbers. Each box contains a random real number, which is the same over all the rooms (that is, box n contains the same real number in every room).

100 set theorists play a game. Each person will go into a unique room and open as many boxes as they like (perhaps countably many) as long as they leave at least one box in their room unopened. Then, each of them need to pick an unopened box in their room, and guess what real number is inside of it.

In order to win, 99 of them need to guess correctly.

The mathematicians can discuss a strategy beforehand, but after they go into their respective rooms, no more communication is allowed. What is a 100% winning strategy for this seemingly impossible task?

The solution is here: https://www.reddit.com/r/math/comments/rpbhos/predicting_random_real_numbers_with_the_axiom_of/?utm_source=share&utm_medium=ios_app&utm_name=iossmf

My question is why can the mathematicians agree on a specific representative for each equivalence class? I thought that the axiom of choice guaranteed a choice function existed, but didnt specify it. But for the solution to the riddle, they would need to know what the choice function is. Can someone explain?

This may be a duplicate, but I’ve done some searching and I can’t find exactly this problem setting, probably due to not knowing the right terminology for how to refer to the transition matrix.

I’m considering absorbing random walks ${X_t}$ which have symmetric transition probabilities inside a window of size $k$. For example, on ${0, 1, 2, dots, 100}$, with $k=3$ if $X_t = 5$
begin{align*}
P[X_{t+1} = 4] &= P[X_{t+1} = 6]\
P[X_{t+1} = 3] &= P[X_{t+1} = 7]\
P[X_{t+1} = 2] &= P[X_{t+1} = 8]\
end{align*}

When the walk is within $k$ of the endpoint, the allowable transitions are truncated on both sides, e.g. if $X_t = 1$, $P[X_t = 3] = 0$.

I wanted to figure out the probability of the walk being absorbed at the upper endpoint, so I did some numerical experiments with walks on the integers from 0-100.
I calculated $T^n$ for the transition matrix $T$ and a large $n$. I then looked at $(T^n)_{i, 100}$ to see the probability of being absorbed at 100. It appears that the value is always equal to $i/100$, regardless of how I set the transition distribution or window-size.

The transition matrix for a walk on the integers in $[0, N]$ has the following properties:

1. $T$ is (row) stochastic
2. $T$ is a band matrix with upper and lower band size $k$
3. $T_{i, j} = 0$ if $|i – j| > max (i, N – i)$
4. $T_{i,i + r} = T_{i, i – r}$ (this is the main property I don’t know the name of)
5. $T_{0, 0} = T_{1, 1} = 1$

I would like a proof or counterexample of the following statement:
begin{equation*}
limlimits_{n to infty} (T^n)_{i, N} = i/N
end{equation*}

If the statement above is true, is it still true if the transition distributions are no longer left-right symmetric, but it’s still the case that $E[X_{t + 1}] = X_t$?

Let $M$ be a von Neumann algebra and $varphi: M_+to [0, infty]$ be a normal, faithful semifinite weight. Consider its associated semi-cyclic representation
$$pi_varphi: Mto B(mathfrak{H}_varphi)$$
(see Takesaki’s second book, chapter VII for details).

Consider the associated map
$$eta_varphi: mathfrak{n}_varphito mathfrak{H}_varphi.$$

Is it true that $eta_varphi(mathfrak{n}_varphicap mathfrak{n}_varphi^*)$ is norm-dense in $mathfrak{H}_varphi$?

Attempt: If $xi perp eta_varphi(mathfrak{n}_varphicap mathfrak{n}_varphi^*)$, then for $x,yin mathfrak{n}_varphi$
$$0 = langle xi, eta_varphi(x^*y) rangle = langle pi_varphi(x)xi, eta_varphi(y)rangle$$
so that $pi_varphi(mathfrak{n}_varphi)xi=0$. Maybe this is sufficient to conclude that $xi= 0$? I feel like non-degeneracy of $pi_varphi$ is relevant.

Let $ W $ be a sub vector space of $ mathbb{R}^n $. How can we determine if $ W $ admits an integer basis?

This is equivalent to asking how to determine if $ W cap mathbb{Z}^n $ spans $ W $.

Let $G$ be a locally compact Polish (or compact) group acting continuously on a locally compact Polish (or compact) space $X$, and $mu$ a Borel measure on $X$. To be sure, continuity of the action means that the map $(g, x) in G times X mapsto g cdot x in X$ is continuous with respect to the product topology on $G times X$. Let $X/G$ denote the orbit space endowed with the quotient topology, and $pi : X rightarrow X/G$ denote the orbit map. A cross-section to the orbit map is a map $s: X/G rightarrow X$ satisfying $s circ pi = 1_{X}$. If $s$ and $t$ are measurable or continuous cross-sections to the orbit map, then it is known that their images $s(X/G)$ and $t(X/G)$ are measurable (closed in the case of a continuous cross-section with compact $G$ and $X$). What is the relationship between $mu(s(X/G)$ and $mu(t(X/G))$? Is it reasonable to expect $mu(s(X/G)) = mu(t(X/G))$?

PS: It is enough for me to consider the case of $X = G$, that is, $X$ is the underlying topological space of $G$, and the action of conjugation, and $mu$ Haar measure on $G$. Thanks.