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 for $C^k$-manifolds with $k<\infty$. 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 $C^\infty$.C^\infty$, whereas$g$would only be of class$C^{k-1}$if$f$were of class$C^k$. 1 The Zariski tangent space at any point of a positive dimensional$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\$.