Neighborhoods in product spaces

Question

Consider the infinite product space where each component is the real numbers with the Euclidean topology and $p = (1,1,1,...)$. Show or disprove that $\forall U \ni p, \exists t > 0, \forall s \in (1, 1+t), (s,s,s,...) \in U$. I'm not really sure where to begin here, any hints would be appreciated, thanks!

Edit: Forgot to clarify I'm using the product topology as opposed to the box.

Answer 1

True, since in the product topology open sets are of the form $\prod_{i\in \Bbb N}U_i$, where $U_i=\Bbb R$ for all but finitely many $i$. Each of the $U_i$ is either of the form $(1-\epsilon_i,1+\epsilon_i)$ or equal to $\Bbb R$.

Let $t=\operatorname{glb}\{\epsilon_i:U_i\ne\Bbb R\}$.

Then consider $(1+t/2,1+t/2,\dots)$.

Answer 2

Hint. A basis open set for the product topology is of the form $\prod_{i\in I\text{ finite}}U_i\times \prod_{i\in \mathbf{N}-I} \mathbf{R}$, where $U_i\subset \mathbf{R}$ is open. An open set in the product topology is the union of basis open sets.

Let $U$ be an open set in the product topology that contains $p$, then $p$ is contained in some basis open set. Can you continue?

Answer 3

Let $U$ be an open neighbourhood of $p$.

Then there exists a basic open neighbourhood of $p$, namely $\prod_n U_n$, such that $p \in \prod_n U_n \subseteq U$, where there is a finite subset $F$ of $\Bbb N$ such that $U_n$ is a proper open subset of $\Bbb R$ (containing $1$) for $n \in F$ and $U_n = \Bbb R$ for $n \notin F$.

For every $n \in F$ pick $r_n >0$ such that $(1-r_n, 1+r_n) \subseteq U_n$, by openness of $U_n$.

Let $t=\min_{n \in F} r_n >0$. (Note that we have a finite set, so the minimum is well-defined and $>0$.)

Then for $s \in (1,1+t)$ we see that $\underline{s}=(s,s,s,\ldots) \in \prod_n U_n $: if $n \in F$, then $1 < s < 1+t \le 1+r_n$, so $s \in (1-r_n,1+r_n) \subseteq U_n$ and if $n \notin F$, $s \in U_n=\Bbb R$ is trivial.

So $t$ is as required.