Prove that the intersection of two neighborhoods of a point…

Prove that the intersection of two neighborhoods of a point $z_0\in\mathbb C$ is again a neighborhood of $z_0$.

Step 1 · Definition

$U$ is a neighborhood of $z_0$ if there exists $\varepsilon>0$ with $\ B(z_0,\varepsilon)=\{z:|z-z_0|<\varepsilon\}\subseteq U$.

Step 2 · Set up the two neighborhoods

Let $U_1,U_2$ be neighborhoods of $z_0$: there exist $\varepsilon_1,\varepsilon_2>0$ with $B(z_0,\varepsilon_1)\subseteq U_1$ and $B(z_0,\varepsilon_2)\subseteq U_2$.

Step 3 · Take the smaller radius

Let $\varepsilon=\min\{\varepsilon_1,\varepsilon_2\}>0$. For any $z\in B(z_0,\varepsilon)$: $\ |z-z_0|<\varepsilon\le\varepsilon_1$, so $z\in B(z_0,\varepsilon_1)\subseteq U_1$; likewise $|z-z_0|<\varepsilon\le\varepsilon_2$, so $z\in B(z_0,\varepsilon_2)\subseteq U_2$.

Step 4 · Conclusion

Every $z\in B(z_0,\varepsilon)$ lies in both $U_1$ and $U_2$, hence in $U_1\cap U_2$:

$$ \boxed{B(z_0,\varepsilon)\subseteq U_1\cap U_2, \qquad \varepsilon=\min\{\varepsilon_1,\varepsilon_2\}>0} $$

so $U_1\cap U_2$ is a neighborhood of $z_0$.

Remark · Finite vs. infinite intersections

The same argument (taking the minimum of finitely many positive radii) shows the intersection of any finite collection of neighborhoods of $z_0$ is again a neighborhood. This fails for infinite collections: taking $U_n=B(0,\frac1n)$ for $n=1,2,3,\ldots$ gives $\bigcap_{n=1}^\infty U_n=\{0\}$, which contains no open disk around $0$ and so is not a neighborhood of $0$ — the infimum of the radii is $0$, not a valid $\varepsilon$.