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 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 2^{*}, and their associated set is the set of all infinite sequences that have that string as prefix.

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

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

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

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

Note that 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 T \cup S open?

If not could you provide a counter example?

Thank you very much,