Why does the fixed point theorem hold for every lambda term? Can someone give a clear and simple answer for why the fixed point theorem holds for every $\lambda$-term, in contrast with the fact that not all numerical function
have a fixed point?
 A: Numerical functions typically accept only numbers as arguments, whereas lambda-terms can accept other lambda-terms.
For your example, $\lambda x.x+1$, the following is a fixed point:
$$ Y(\lambda x.x+1)$$
where $Y$ is Turing's Y-combinator: $Y=UU$ and $U=\lambda ux.x(uux)$
This is a fixed point because:
$Y(\lambda x.x+1) = UU(\lambda x.x+1) = (\lambda ux.x(uux))U(\lambda x.x+1) = (\lambda x.x+1)(UU(\lambda x.x+1)) = (\lambda x.x+1)(Y(\lambda x.x+1))$
(using '=' loosely for congruence or $\beta$-reduction)
We have shown $(\lambda x.x+1)(Y(\lambda x.x+1))$ is $\beta$-equivalent to $Y(\lambda x.x+1)$, meaning that $Y(\lambda x.x+1)$ is a fixed point of $\lambda x.x+1$
Turing's Y-combinator will actually find a fixed point for any lambda-term $X$. That is, following the steps above but generalizing the example $\lambda x.x+1$ to any $X$, we get $X(YX)=_\beta YX$, so $YX$ is a fixed point of $X$. This is essentially the proof of the fixed point theorem.
$Y(\lambda x.x+1)$ is a lambda-term, so it's not a numerical fixed point of $x \rightarrow x+1$
A: As was noted by others lambda terms are more general than numerical functions so you are able to draw answer(fixed point) from different set. While for numerical function e.g. $f: \mathbb{R} \rightarrow \mathbb{R}$ your fixed point must belong to $\mathbb{R}$. On contrary lambda terms reduces to other lambda terms some of them can designate numbers from $\mathbb{R}$ some of them do not designate the numbers.
For more rigorous explanations you should read some proofs of fixed point theorem for lambda calculus.
