Let X be the set of all real sequences. For both the box and the product topology, find the closure and the interior of S, where S is the subset of X containing: (a) bounded sequences (b) sequences avoidind a fixed finite set F (c) sequences assuming only a finite set F of values (d) strictly increasing sequences
Now, I think the Prof. swapped the first and the second column in his solution, and the first inclusion between interiors, because if topology B is coarser than topology T then both the interior and the closure (of S) in B are contained in those in T.
My solution is, β-int, τ-int ------- β-cl, τ-cl a) ∅, S ------- S, S b) ∅, S ------- X, X c) ∅, ∅ ------- S, S d) ∅, S ------- L, L
where L is the set of non-decreasing sequences. So, am I mistaken? Looking at the difference in the solutions, I reasoned like this: Every non-bounded sequence has a whole τ-neighbourhood of their kind, so S is closed.
The box topology is the finer topology so will have possibly larger interior and possibly smaller (!) closure.
For bounded sequences: this has empty product interior because any basic open set has unbounded sequences (we only "control" finitely many coordinates). But the box neighbourhood $(x_1 -1 ,x_1 + 1) \times \ldots \times (x_n - 1, x_n +1) \times \ldots$ around a bounded sequence $(x_n)$ only contains bounded sequences, so then $S$ is open and equal to its own interior. Also that same neighbourhood around an unbounded sequence only contains unbounded sequences, so $S$ is box-closed. But as said, all basic product open sets contain unbounded sequences, so $\overline{S} = X$ for the product topology. So your profs first row is completely correct: (box interior, box closure) $S$, $S$, (product interior, product closure) $\emptyset$, $X$.
Sequences avoiding a fixed finite set $F$: the product topology: if $(x_n)$ is a sequence in $S$, we cannot control it at finitely many coordinates and be garanteed to have a sequence in $S$ again, so $\operatorname{int}_\tau(S) = \emptyset$. In the box topology we see that $S = \prod_n (\mathbb{R}\setminus F)$ which is box open so $\operatorname{int}_\beta(S) = S$. If $x$ is a sequence not in $S$, meaning $x_n \in F$ for some $n$, any box-open set around $x$ also will contain points of $S$ too, as some thought shows. So $\overline{A}^{(\beta)} = X$, so certainly the same is true for the product topology. Second row also confirmed.
c) Sequences with finite range: empty product interior, empty box interior, dense in product topology (if $O$ is basic open, we pick a sequence with some values at the non-trivial opens and $0$ everywhere else, so that $S$ intersects all product open sets), but closed in the box topology (a sequence with infinite range $x_{n_1},\ldots x_{n_k},$ has pairwise disjoint open intervals around each of these values, and so any other sequence in the box neighbourhood built that way has infinite range too), so $\emptyset$, $\emptyset$, $X$, $S$. So one difference with your profs answer and your answer is wrong too. $S$ is product-dense not closed.
(d) stricly increasing sequences: box open by using disjoint intervals around the $x_n$ again. Empty interior for product: no control. The non-increasingness is a condition on just two coordinates, so $S$ is product closed and thus also box closed. The last row should be $S\, \emptyset\,\, S\, S$