What is the homotopy type of the affine space in the Zariski topology..? I'm asking this question out of curiosity, as I was unable to come to a conclusion. 
Consider the affine space over an algebraically closed field, for concreteness we can work with $\mathbb{C}^{n}$. Let's restrict ourselves to the closed points, ie. we're working with the spectrum of maximal ideals. What is the homotopy type of this space..? 
I'd be happy with only knowing the weak homotopy type and I have some suspicion that $\mathbb{C}^{n}$ might be weakly contractible, but I was unable to prove it. Since the Zariski topology $\mathbb{C}^{n}$ is strictly weaker than the usual one, it is of course easy to construct homotopies between maps $S^{n} \rightarrow \mathbb{C}^{n}$, as all the homotopies in the usual topology will work. However, the problem lies in the fact that, in principle, there might be simply much more continous maps from the sphere. 
One might also try to consider the "radial contraction" map $r: \mathbb{C}^{n} \times \mathbb{C} \rightarrow \mathbb{C}^{n}$ given by $r(v, t) = tv$ which is also continous in the Zariski topology and then restrict that to $\mathbb{C}^{n} \times [0, 1]$ to prove contractibility directly. However, Zariski topology on the product is stronger than the product topology, so it doesn't work. 
Edit: I have considered the case of $\mathbb{C}$ and it seems it is indeed contractible, but maybe I'm missing something obvious. We're looking for a map $H: \mathbb{C}^{1} \times I \rightarrow \mathbb{C}^{1}$ that restricts to the identity on $\mathbb{C}^{1} \times \{ 1 \}$ and to the constant map on $\mathbb{C}^{1} \times \{ 0 \}$. Note that it's enough that the preimages of points are closed, since the closed sets in $\mathbb{C}^{1}$ are finite or the whole space. Choose a bijection $\phi: \mathbb{C}^{1} \times (0, 1) \simeq \mathbb{C}^{1}$ and put $H(z, t) = \phi(z, t)$ for $t \in (0, 1)$. Then the preimage of every in $\mathbb{C}^{1}$ under $H$ is some closed subset of $\mathbb{C}^{1} \times \{ 0, 1 \}$ plus a single point, so it is closed. 
 A: Here is a proof that for every $n\ge 2$, the affine space ${\mathbb C}^n$ (with Zariski topology) is contractible. (Since you already know how to do it for $n=1$, I am treating only the case $n\ge 2$.) Sadly, this proof reveals nothing interesting in algebra/algebraic geometry. The same proof works for any variety over ${\mathbb C}$. 
Let $\Delta_R$ denote the open disk of radius $R$ in ${\mathbb C}$, centered at $0$. 
I will need:
Lemma. For every $m\ge 1$ there exists a holomorphic function $F: \Delta_1\to {\mathbb C}^{m}$ whose graph $\Gamma=\Gamma_F$ satisfies the property that for every proper affine subvariety $V\subset {\mathbb C}^{m+1}$, the intersection $V\cap \Gamma$ is finite. 
Proof. Let $f_1,...,f_m: \Delta_2\to {\mathbb C}$ be algebraically independent holomorphic functions. Let $F$ denote the restriction of the function $f=(f_1,...,f_m)$ to the disk $\Delta_1$. The graph of $F$ satisfies the required property. Indeed, algebraic independence of the functions $f_1,...,f_n$ implies that the intersection $V\cap \Gamma_f$ has to be zero-dimensional. But this intersection is an analytic subvariety; hence, its intersection with the tube $\overline{\Delta_1} \times {\mathbb C}^n$ is finite. qed   
Now, consider the following map $H: {\mathbb C}^n\times {\mathbb C}\to {\mathbb C}^n$:


*

*$H(z_1,...,z_n,1)=(z_1,...,z_n)$. 

*$H(z_1,...,z_n, 0)=(0,...,0)$. 

*Consider the subset $E\subset {\mathbb C}^{n+1}$ consisting of tuples $(z_1,...,z_n,w)$ such that $w\notin \{0, 1\}$. Since the graph $\Gamma$ as above has cardinality continuum, there exists a bijection $H|_E: E\to \Gamma\subset {\mathbb C}^n$. 
I claim that  the function $H$ thus obtained is continuous in Zariski topology (on domain and the range). Let $V\subset {\mathbb C}^n$ be a Zariski closed subset (an affine subvariety). Suppose first that $V$ does not contain $0\in {\mathbb C}^n$. Since the intersection $V\cap \Gamma$ is finite (and $H$ restricted to the complement of ${\mathbb C}^n \times 0$ is the identity map), $H^{-1}(V)$ is the union of a finite set and $V\times 0$. Such set is clearly Zariski closed. The case when $V$ contains $0$ is similar, you just have to add the subvariety ${\mathbb C}^n \times 0$ to the above inverse image. Therefore, $H$ is continuous. qed
A: Here is a more elementary version of Moishe Cohen's argument, which also works over more general fields.  Let $K$ be any field of cardinality $\geq 2^{\aleph_0}$; we will show that $K^n$ is contractible in the Zariski topology.  Let $B\subset K$ be an algebraically independent set of cardinality $|K|$ and partition $B$ into sets of $n$ elements.  Use these sets of $n$ elements as the coordinates of points of $K^n$, giving a set $S\subset K^n$ of cardinality $|K|$.  I claim that if $f\in K[x_1,\dots,x_n]$ is a nonzero polynomial, it has only finitely many zeroes in $S$.  Indeed, if $f(s)=0$, that means the coordinates of $s$ are algebraically dependent over the field generated by the coefficients of $f$, and this can only happen for finitely many $s\in S$.
Now, as in Moishe Cohen's answer, we can define a contraction $H:K^n\times[0,1]\to K^n$ by $H(x,0)=x$, $H(x,1)=0$, and on $K^n\times (0,1)$, $H$ is given by some bijection $K^n\times(0,1)\to S$ (here is where we use that $|K|\geq 2^{\aleph_0}$).
In fact, we can go even further.  Assume $K$ is algebraically closed, and let $X$ be any irreducible variety over $K$.  Let $Y\subseteq X$ be a dense affine open subvariety, and by Noether normalization let $f:Y\to K^n$ be a finite surjective morphism.  The set $f^{-1}(S)\subset X$ then has the property that its intersection with any Zariski-closed proper subset of $X$ is finite (here is where we use irreducibility of $X$, to guarantee that there is no closed proper subset of $Y$ on which $f$ is surjective).  We can then again construct a contraction $H:X\times [0,1]\to X$ as above, with $f^{-1}(S)$ in place of $S$.  So any irreducible variety over an algebraically closed field of cardinality $\geq 2^{\aleph_0}$ is contractible in the Zariski topology.
