Flip til you get heads; How is the event that you never get heads part of the sample space? Experiment: flip a coin until you get Heads. I would like to understand better why $E=(T,T,T,....)$, the event where heads is never flipped, is part of the sample space. 
Isn't the experiment going on forever unfinished in that case? Whereas if someone asked if $(T,T,T)$ were an element of the sample space, the answer would be no, because elements of the sample space must look like $(T,T,...,T, H)$ or  by exactly $E$. 
To clarify, is $E$ an element of the sample space of this experiment (I claim it is). If so, can you explain why?
Thanks.
 A: There is in such cases always a part that is definitively a mathematical part, and a part that is definitively a modeling part.


*

*The mathematical part starts when we already have a probability space $(\Omega,\mathcal F,\Bbb P)$ and ask something related to probability theory, e.g. for a given event in $\mathcal F$ which is its probability.

*The modeling part is the part from a "phenomenological story" till we have the above mathematical setting, and we "consider by model" that the space reflects exactly what we need.
In our case, the OP addresses only the modeling part. The space has than two choices.
First choice: The space $\Omega$ is a set with elements $(H)$, $(TH)$, $(TTH)$, ... , $(T^nH):=\underbrace{TT\dots T}_{n\text{ times}}H$, ... and also an element denoted by $T^\infty = (TTT\dots)$. (So far we have only notations.) The above elements can be also denoted by $0,1,2,\dots, n,.\dots$ together with an element $\infty$.
Now we have to declare a $\sigma$-algebra. We take it to be the power set of $\Omega$.
Now we finally have to declare the probability. The $\sigma$-algebra is atomic, so we have to declare it on atoms, on $1$-element sets. Our belief about our coin (that it is fair and never gets erosion effects after many $T$'s) lets us give to the alternatively denoted elements $0,1,2,\dots,n,\dots$ (i.e. to the corresponding 1-element sets) the probabilities 
$$
\frac 12,\ 
\frac 14,\ 
\frac 18,\ 
 \dots ,\ 
\frac 1{2^n},\ \dots
$$
so that we have to set $\Bbb P(\{\infty\})=0$. (Especially when discussing with physics experts, they tend to think about a very special situation, and feel in it an electromagnetic field around the coin, so that this one probability is in fact felt like $\Bbb P(\{\infty\})=9/10$ after neglecting quantum effects, and we have to redistribute the rest to $0,1,2,\dots$ - and there is no chance to convince them that this makes no sense. It is just a matter of modeling the matter.)
Second choice: As above, but we restrict to the subspace $\Omega_0=\Omega-\{\infty\}$ with the induced probability. There is no $(TTT\dots)=\infty$ in it. And the inclusion omits in the image a zero set. So we have probabilistically the "same" situation. 
(Similar situation, when we Lebesgue-integrate from $0$ to $1$ on the real line, do we integrate on the space $[0,1]$ or on the space $(0,1)$?) 

For short, you have a mathematical question only when you give the space, but the question was not giving any space.
