The Zariski tangent space at any point of a positive dimensional  $C^1$-manifold $X$ has dimension $2^{2^{\aleph_0}}= 2^{\frak c}$.
Let me explain in the case when $X=\mathbb R$.

Consider the ring $C^1_0$ of germs of $C^1$- functions at $0\in \mathbb R$ and its maximal ideal $\frak m $ of germs of functions vanishing at zero. The cotangent space at zero of $\mathbb R $ is $Cot_0=\frak m /\frak m ^2$ and the Zariski tangent space is $T_0=(Cot_0)^{\ast}$ (dual $\mathbb R$-vector space).
Now the germs of the functions $x^\alpha $ are linearly independent modulo $\frak m ^2$ for $\; \alpha\in(1,2)$ . Hence $dim_{\mathbb R} (Cot_0)=\frak c$ and so indeed the Zariski tangent space at zero of $\mathbb R$ is $dim_{\mathbb R} (T_0)=2^{\frak c}$.

It is noteworthy that many textbooks erroneously claim that for an $n$-dimensional manifold of class $C^1$ the Zariski tangent space defined above has dimension $n$. Or they make some equivalent mistake like claiming that the vector space of derivations of $C^1_0$ has dimension $n$ . An example of such an error is on page 42 in Claire Voisin's (excellent!) book  *Hodge Theory And Complex Algebraic Geometry I* published by Cambridge University Press.

To end on a positive note, the phenomenon I am describing only raises 
its ugly head if $k<\infty$. For $n$-dimensional $C^\infty$-manifolds the Zariski tangent space at any point has dimension $n$, as it should. The heart of the matter is that a $C^\infty$ function $f$ , 
on $\mathbb R$ say, which vanishes at zero can be written $f=xg$ for some function $g$ which is also of class $C^\infty$.