Dimension of product of irreducible topological spaces Let X and Y are irreducible spaces, do we have 
$$\dim (X\times Y) = \dim X + \dim Y ?$$
We know that it is true for irreducible varieties, but here $X$ and $Y$ are only topological spaces. 
 A: In this answer, I'll use the definition of dimension given in the comments: $\dim(X)$ is the supremum of integers $n$ such that there exists a chain of length $n+1$ of irreducible closed subsets of $X$: $A_0\subsetneq A_1\subsetneq\dots \subsetneq A_n$. If there are arbitrarily long such chains, then $\dim(X) = \infty$. 
With this definition, $\dim(X\times Y) = \dim(X) + \dim(Y)$ is true for any nonempty topological spaces $X$ and $Y$. To prove this, we'll first characterize the irreducible closed sets in the product topology.
Lemma: A set $Z\subseteq X\times Y$ is closed and irreducible in the product topology if and only if it has the form $A\times B$, where $A$ and $B$ are closed irreducible subsets of $X\times Y$. 
Proof: In one direction, suppose $A\subseteq X$ and $B\subseteq Y$ are irreducible closed sets. Then $A\times B$ is closed - it is the compliment of the open set $(A^c\times Y)\cup (X\times B^c)$. We want to show it's irreducible. I think this is easiest to see if we use the formulation that a closed set $C$ is irreducible if and only for any open sets $U_1$ and $U_2$, if $U_1\cap C$ and $U_2\cap C$ are nonempty, then $U_1\cap U_2\cap C$ is nonempty. In fact, it suffices to check this condition for basic open sets, which in the product topology can be written $U_1\times V_1$ and $U_2\times V_2$. If $(U_i\times V_i)\cap (A\times B)$ is nonempty for $i = 1$ and $2$, then $U_i\cap A$ is nonempty for $i = 1$ and $2$, so there is some $x\in U_1\cap U_2\cap A$, since $A$ is irreducible. Similarly, there is some $y\in V_1\cap V_2\cap A$. But then $(x,y)\in (U_1\times V_1) \cap (U_2\times V_2)\cap (A\times B)$.
In the other direction, suppose $Z\subseteq X\times Y$ is closed and irreducible. Let $A = \overline{\pi_X(Z)}$ (the closure of the projection of $Z$ to $X$), and let $B = \overline{\pi_Y(Z)}$. Passing to the subspaces $A$ and $B$, we have that $Z\subseteq A\times B$ is closed and irreducible. Suppose for contradiction that it is a proper subset. Then there is some basic open set $U\times V\subseteq Z^c = (A\times B)\setminus Z$, with $U\subseteq A$ and $V\subseteq B$ open. And since $Z\subseteq (U^c\times B)\cup (A\times V^c)$ is irreducible, $Z$ is contained in one of these sets. If $Z\subseteq U^c\times B$, then $\pi_X(Z)\subseteq U^c$, contradicting the fact that $A = \overline{\pi_X(Z)}$, and similarly if $Z\subseteq A\times V^c$. $\square$
On to dimensions. Suppose we have a chain $A_0\subsetneq A_1\subsetneq\dots \subsetneq A_n$ witnessing $\dim(X)\geq n$ and a chain $B_0\subsetneq B_1\subsetneq\dots \subsetneq B_m$ witnessing $\dim(Y)\geq m$. Then the chain $$A_0\times B_0\subsetneq A_1\times B_0\subsetneq \dots\subsetneq A_n\times B_0\subsetneq A_n\times B_1\subsetneq \dots \subsetneq A_n\times B_m$$
witnesses $\dim(X\times Y)\geq n+m$. Indeed, all these sets are irreducible and closed, by the lemma. This shows $\dim(X\times Y)\geq \dim(X)+\dim(Y)$. 
We'll prove the other inequality, $\dim(X\times Y)\leq \dim(X)+\dim(Y)$, by induction. The base case, $\dim(X\times Y) = 0$, is clear: the closure of a point is always irreducible, so any nonempty set has dimension at least $0$. Now suppose we have a chain $Z_0\subsetneq \dots \subsetneq Z_{n+1}$ witnessing $\dim(X\times Y) = n+1$. Then the subspace $Z_n$ has dimension $n$, and $Z_n = A\times B$ for $A\subseteq X$ and $B\subseteq Y$ closed and irreducible, by the lemma. By induction, $\dim(A) + \dim(B) \geq n$, witnessed by chains $A_0\subsetneq \dots\subsetneq A_i = A$ and $B_0\subsetneq \dots \subsetneq B_j = B$, with $i+j\geq n$. But we can also write $Z_{n+1} = A'\times B'$ for $A'$ and $B'$ closed irreducible, and since $Z_n\subsetneq Z_{n+1}$, we must have $A\subsetneq A'$ or $B\subseteq B'$. Hence at least one of the two chains (in $X$ or in $Y$) can be extended by at least one, and $\dim(X) + \dim(Y) \geq i+j + 1 \geq n+1$.
A final note: The point of this argument is that we don't get any interesting new irreducible closed sets in the product topology. Since the question mentions varieties, it's important to remember that if $X$ and $Y$ are algebraic varieties with the Zariski topology, and $X\times Y$ is the product of varieties, then the Zariski topology on $X\times Y$ is not the product topology. In fact, we get lots of new irreducible closed sets, e.g. all the plane curves in $\mathbb{A}^2 = \mathbb{A}^1 \times \mathbb{A}^1$ which aren't just lines parallel to the axes.
A: Sorry, I don't have enough reputation to make this a comment. Are you sure that $\mathop{dim}(X \times Y)=\mathop{dim} X+\mathop{dim}Y$ even for irreducible varieties? This is certainly true if we take the product as varieties, but the topology on that product variety will typically not agree with the product topology.
