I have this question about topology, but I'm really not an expert, so please bear with me if I use some not very common notations (Please ask for any clearification!)

Consider the cantor space $\displaystyle 2^{\omega}$, i.e., the space of all infinite sequences of 0s and 1s.
Consider its standard topology: the basic open sets are (representable) by finite strings in $\displaystyle 2^{*}$, and their associated set is the set of all infinite sequences that have that string as prefix.

I use the following notation: given $\displaystyle x\in 2^{\omega}$ and $\displaystyle n\in \mathbb{N}$, I write $\displaystyle x\downarrow n$ to denote the string consisting of the first n bits of the infinite sequence x.

Consider the Borel $\displaystyle \sigma$ algebra generated by the basic sets.

Let $\displaystyle S$ be a Borel set.
Now suppose there is a Borel set $\displaystyle T$ such that

  1. T is an open set
  2. For every $\displaystyle x\in S$, and for every natural number $\displaystyle n$, there exist an $\displaystyle y\in T$ such that $\displaystyle x\downarrow n = y \downarrow n$.

Note that $\displaystyle T\not \subseteq S$.
I believe the point 2) means that (Please correct me if i'm wrong):

  1. T is dense in S
  2. S is included in the clousure of T.

My question is the following:

is $\displaystyle T \cup S$ open?

If not could you provide a counter example?

Thank you very much,