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We just computed in class a few days ago that $$H^*(\mathbb{R}P^n,\mathbb{F}_2)\cong\mathbb{F}_2[x]/(x^{n+1}),$$ and it was mentioned that $H^*(\mathbb{R}P^\infty,\mathbb{F}_2)\cong \mathbb{F}_2[x]$, but I (optimistically?) assumed that the cohomology ring functor would turn limits into colimits, and so $$\mathbb{R}P^\infty=\lim\limits_{\longrightarrow}\;\mathbb{R}P^n$$ would mean that we'd get the formal power series ring $$\lim\limits_{\longleftarrow}\;\mathbb{F}_2[x]/(x^{n+1})=\mathbb{F}_2[[x]].$$

My professor said that the reason we get $\mathbb{F}_2[x]$ is just that the cohomology ring, being the direct sum of the cohomology groups, can't have non-zero elements in every degree, which certainly makes sense.

Even though that should settle the matter, for some reason, I'm still having a bit of trouble making this make click for me. Is the explanation just that the cohomology ring functor doesn't act as nicely as I'd hoped? Is there an intuitive explanation of what's going on?

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    I'll wait for a real expert to leave a precise answer, but...it seems that your intuition about the cohomology ring functor needs a little calibrating. How about this: you have more than just a ring structure on $H^*$ -- you have a **graded algebra structure**. Really you might as well view this as a sequence of functors into $R$-Mod *together* with a graded-commutative bilinear product on them. In particular though the property that each graded piece is finitely generated is fundamental and should be viewed as *nice*, I think....2012-02-20
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    ...But I don't doubt that the *completion* of the cohomology ring functor appears elsewhere in algebraic topology. That's really what I'm waiting for an expert to weigh in on.2012-02-20
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    (Above, when I say that each graded piece is finite-dimensional, I don't mean that this holds for $H^*(X,R)$ for *all* spaces $X$, but rather that it holds for sufficiently nice spaces, namely those homotopy equivalent to a CW-complex with finite $n$-skeleta for each fixed $n$.)2012-02-20
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    I do not know if this helps, but this was proved explicitly by Hatcher in his book, page 214. By what you wrote I assume you already know the proof, though.2012-02-20
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    Since $RP^\infty$ has a model with finitely many simplices in each dimension, its cohomology is of at most of countable dimension over the base field.2012-02-20
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    @Pete: Ah, I see - the inverse limit of $\mathbb{F}_2[x]/(x^n)$ as *graded* $\mathbb{F}_2$-algebras is just $\mathbb{F}_2[x]$; whereas $\mathbb{F}_2[[x]]$ does not have a grading compatible with the maps to each $\mathbb{F}_2[x]/(x^n)$.2012-02-20
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    @PeteL.Clark: The graded completion of the cohomology ring does appear in algebraic topology. For example the L-class appearing in Hirzebruch's signature theorem can be regarded as an element of the graded completion of the cohomology ring.2012-02-20

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