Under what conditions on the sequence {$a_n$} (all $a_n$ > 0) the set parallelepiped {$x \in l_p: x_n < a_n$} is an open subset of $l_p$, where $p \in [1, +\infty)$?

1$\begingroup$ the answer is $inf {a_n}$ > 0 $\endgroup$ – Mr White yesterday

$\begingroup$ One may also note that for $p=+\infty$ the analog set $\{x \in l_\infty: x_n < a_n, \forall n \}$ is never open. $\endgroup$ – Pietro Majer yesterday

$\begingroup$ $l_p$ of what space, $l_p(\mathbb{N})$? $\endgroup$ – Amir Sagiv yesterday

$\begingroup$ @PietroMajer : I think that, if e.g. $a_n=n$ for all $n$, then your set is open in $l_\infty$. $\endgroup$ – Iosif Pinelis yesterday

$\begingroup$ @PietroMajer : I have now given an iff condition for $p=\infty$ as well. $\endgroup$ – Iosif Pinelis yesterday
The parallelepiped $P:=\{x\in l_p\colon x_n<a_n\ \forall n\}$ is open iff $a:=\inf_n a_n>0$.
Indeed, suppose that $P$ is open. Since $0\in P$, we have $B_0(r)\subseteq P$ for some real $r>0$, where $B_z(r)$ denotes the open ball of radius $r$ centered at $z$. For each natural $k$, the point $y^{(k)}$ with coordinates $y^{(k)}_n=(r/2)1(n=k)$ is in $B_0(r)\subseteq P$ and hence $r/2<a_k$, for all natural $k$. So, $a>0$.
Vice versa, suppose that $a>0$. Take any $x\in P$. Take any natural $N$ such that $\sum_{n>N}x_n^p<(a/2)^p$. Take any $r>0$ such that $r<a/2$ and $r<a_nx_n$ for all natural $n\le N$. Then $B_x(r)\subseteq P$. Indeed, take any $y\in B_x(r)$. Then $y_n\lex_n+r<a_n$ for $n\le N$. Also, for any $n>N$ we have $$x_n\le\Big(\sum_{m>N}x_m^p\Big)^{1/p}<a/2,$$ whence $y_n\lex_n+r<a/2+a/2=a\le a_n$. So, $y\in P$.
So, $P$ is open.
As a free bonus, consider the case $p=\infty$. Then $P$ is open iff $a_n\to\infty$.
Indeed, suppose that $P$ is open while $b:=\liminf_na_n<\infty$. Let $x_n:=\min(b,(11/n)a_n)$. Take any real $r>0$ and let $y_n:=x_n+r/2$ for all $n$. Then $x\in P$ and $y\in B_x(r)$. Moreover, $\liminf_n y_n=b+r/2>b$, so that $y\notin P$, which contradicts the assumption that $P$ is open. Thus, if $P$ is open, then $a_n\to\infty$.
Vice versa, suppose that $a_n\to\infty$. Take any $x\in P$. Then $x_n<\min(a_n,A)$ for some real $A>0$ and all $n$. Take any natural $N$ such that $a_n>A+1$ for all $n>N$. Take any $r>0$ such that $r<1$ and $r<a_nx_n$ for all natural $n\le N$. Then $B_x(r)\subseteq P$. Indeed, take any $y\in B_x(r)$. Then $y_n\lex_n+r<a_n$ for $n\le N$. Also, for any $n>N$ we have $y_n\lex_n+r<A+r<A+1<a_n$. So, $y\in P$.
So, $P$ is open.

$\begingroup$ 𝑦(𝑘)𝑛=(𝑟/2)1(𝑛=𝑘) I'm not sure if I understand this correctly $\endgroup$ – Mr White yesterday

1$\begingroup$ I think it's probably better not to answer "I have no idea" questions that look like homework. $\endgroup$ – LSpice yesterday

$\begingroup$ @MrWhite : $1(n=k)$ means $1$ if $n=k$ and $0$ otherwise. So, $y^{(k)}_n=(r/2)1(n=k)$ means $r/2$ if $n=k$ and $0$ otherwise. $\endgroup$ – Iosif Pinelis yesterday

1$\begingroup$ @LSpice : I agree. At first, it didn't occur to me that this may be an exercise, and it took me a bit to figure out this answer, which I think might be not too far from an average MO answer in terms of the difficulty and the length. Only after that it occurred to me that this may be an exercise. $\endgroup$ – Iosif Pinelis yesterday
