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I'm trying to prove that the dimension of the projective space $\mathbb P^n$ is $n$. I've seeing some books saying that since the $\{U_i\}$ ($U_i$ homeomorphic to $\mathbb A^n$) is an open cover of $\mathbb P^n$, we have $\dim \mathbb P^n=\sup\dim U_i=\dim \mathbb A^n=n$.

I didn't understand what the open cover of a topological space has to do with its dimension.

I need a clarification at this point.

Thanks a lot.

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Since $\mathbb P^n$ has $U_i$ as a dense open subset and $U_i\cong \mathbb A^n$ has dimension $n$, we conclude that $\mathbb P^n$ also has dimension $n$.
Right?
Completely wrong !

For example if $X=\text{Spec}(R)=\{\eta ,\mathfrak m\}$ is the spectrum of a discrete valuation ring $(R,\frak m)$ (where $\eta\in X$ is the generic point corresponding to the zero ideal of $R$), the singleton set $U=\{\eta\}\subset X$ is open and dense in $X$.
However $U$ has dimension $0$ (like all singleton spaces !) whereas $X$ has dimension $1$ , so that the "argument" at the beginning of this post is false.

I wrote this "answer" as a warning to over-optimistic beginning algebraic geometers.
A correct proof can be found here, as already mentioned in a comment to the question.

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