Consider the following topology on $\Bbb{R}$: $U$ is open if and only if $\Bbb{R}-U$ is countable or all of $\Bbb{R}$. I claim that if $A$ is an uncountable set in $\Bbb{R}$, then its closure is $\Bbb{R}$ (i.e., it is dense). Recall that the closure of $A$ is the smallest closed set containing $A$, and that closed sets are formed by taking complements of open sets, which obviously makes proper closed subsets countable.
Clearly, then, there is no way a proper closed subset could contain $A$, since an uncountable set cannot be a subset of a countable set, Hence, the closed subset must not be …
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Suppose
$$z\frac{\cos z}{\sin z}= \sum_{-\infty}^{n} a_nz^n $$
the laurent series of $f(z)= z\frac{cosz}{sinz} $ on the ring π<|z|<2π.Find the $a_n$.
Now i know $a_n= \frac{1}{2πi} \int\frac{f(z)}{z^{n+1}}dz$ so for $$n=0$$i plug in the $f$ and i try to use the residues theorem but i dont ...
