Counterexample to Kunneth Formula I'm reading Example 9.14 of Bott&Tu's Differential Forms in Algebraic Topology, which gave a counter example to the Kunneth formula when the assumption is not satisfied by constructing manifolds $M$ and $F$ such that $H^0(M\times F)$ is naturally isomorphic to $\mathbb{R}^{\mathbb{Z}\times\mathbb{Z}}$ while $H^0(M)\otimes H^0(F)$ is naturally isomrophic to $\mathbb{R}^{\mathbb{Z}}\otimes \mathbb{R}^{\mathbb{Z}}$.
I can see that these two are not naturally isomorphic via the explicit mapping in Kunneth formula. But if one ommit the explicit mapping in the statement of Kunneth formula, then a counterexample in degree $0$ needs $H^0(M\times F) \not\cong H^0(M)\otimes H^0(F)$ as real vector spaces. So I wish to figure out if $\mathbb{R}^{\mathbb{Z}\times\mathbb{Z}}$ is isomorphic to $\mathbb{R}^{\mathbb{Z}}\otimes \mathbb{R}^{\mathbb{Z}}$ as vector spaces. The counterexample would appear to be more perfect to me if they are not isomorphic.
Two vector spaces are isomorphic if and only if their basis have a same cardinality. Since there is a natural embedding $\mathbb{R}^{\mathbb{Z}}\otimes \mathbb{R}^{\mathbb{Z}}\hookrightarrow \mathbb{R}^{\mathbb{Z}\times\mathbb{Z}}$, the cardinality of basis of $\mathbb{R}^{\mathbb{Z}}\otimes \mathbb{R}^{\mathbb{Z}}$ is no larger than $\mathbb{R}^{\mathbb{Z}\times\mathbb{Z}}$. On the other hand, we have $\mathbb{R}^\mathbb{Z\times Z}\cong \mathbb{R}^\mathbb{Z}$ by re-indexing $\mathbb{Z}\times \mathbb{Z}\cong\mathbb{Z}$, and then $\mathbb{R}^\mathbb{Z}\hookrightarrow \mathbb{R}^{\mathbb{Z}}\otimes \mathbb{R}^{\mathbb{Z}}:v\mapsto v\otimes 1$ is an embedding, concluding that $\mathbb{R}^{\mathbb{Z\times Z}}\cong \mathbb{R}^{\mathbb{Z}}\otimes \mathbb{R}^{\mathbb{Z}}$.
Is my above argument correct? If so, is there any other counterexample to Kunneth formula, with $H^*(M\times F)\not\cong H^*(M)\otimes H^*(F)$ as real vector spaces in some degree?
Thanks in advance.
 A: I believe there is no such example. But please read carefully and correct me if I am wrong.
First of all the Künneth formula in homology with coefficients in a field $k$ holds for any pair of topological spaces, so
\begin{align*}
H_*(X;k)\otimes H_*(Y;k) \overset{\sim}{\to} H_*(X\times Y;k).
\end{align*}
By the universal coefficient theorem one has $H^*(X;k) \overset{\sim}{\to} H_*(X;k)^* $. So we have isomorphisms
\begin{align*}
H^*(X\times Y;k) \overset{\sim}{\to} (H_*(X\times Y;k))^* \overset{\sim}{\to} \left(H_*(X;k)\otimes H_*(Y;k)\right)^*
\end{align*}
Let $V = H_*(X;k)$ and $W = H_*(X;k)$. If one of those vector spaces is finite dimensional the inclusion \begin{align*} 
V^* \otimes W^* \to (V\otimes W)^*\end{align*} becomes an isomorphism. Composing its inverse with the upper line gives back the cohomology Künneth isomorphism. Now if $V$ and $W$ are both infinite dimensional $V^*\otimes W^* \overset{\sim}{=} (V\otimes W)^*$ still holds (see e.g. this question) but the inclusion $V^* \otimes W^* \to (V\otimes W)^*$ is not an isomorphism. Anyway $H^*(X\times Y;k)$ always has the same dimension as $H^*(X;k) \otimes H^*(Y;k)$.
