What will happen if a sequence of rectangles with decresing diameter tends to a point? I am studying complex analysis, particularly different versions of Cauchy's theorem. Now while going through the proof of Cauchy's theorem for rectangles I found a concept that a sequence of rectangles with decreasing diameter tends to a point of the initial rectangle I have started with. I know the the concept of sequence in metric spaces where sequence of points of the space tends to some point in that space or some extended space. But how can I relate this concept to above when some nested sets with decreasing diameter tends to a point. What I have thought about it is as follows $:$
Suppose $\{R_n \}$ be nested sequence of rectangles in the complex plane with $\mathrm{diam} (R_n) \rightarrow 0$ as $n \rightarrow \infty$. Then since $(\mathbb C, ||\ .||_2)$ is complete so by Cantor's intersection theorem we have $\cap_{n=1}^{\infty} R_n = \{z^* \}$.  I think it means to say that for a given $\epsilon > 0$, $\exists$ $k \in \mathbb N$ such that $\forall n \geq k$ we have $||z-z^*|| < \epsilon$ for all $z \in R_n$. 
Is it the correct interpretation of such a convergence? I am not sure of that. Please check it.
Thank you in advance.
 A: Cantor's intersection theorem applies, not because $\mathbb C$ is complete, but because our rectangles $R_n$ are compact. [See how the compactness property is used in the proof here.]
Actually, Cantor's intersection theorem only tells us that $\bigcap_n R_n$ is non-empty. We want to show that $\bigcap_n R_n$ contains a single point. This is easy to show, and we can do this by following the logic in the third-to-last sentence in your question. Suppose that $\bigcap_n R_n$ contains two distinct points, $z$ and $z'$, and suppose $| z - z'| = \epsilon$, where $\epsilon > 0$. Since $\lim_{n \to \infty}{\rm diam \ }R_n = 0$, there exists an $N$ such that ${\rm diam \ } R_N < \epsilon / 2$. This is a contradiction, because $z, z' \in R_N$ and $|z - z'| = \epsilon > \epsilon / 2 = {\rm diam \ }R_N$.
Having established that $\cap_n R_n$ is a single point, say $\{ z^\star \}$, it is easy to prove that your interpretation of the statement is correct:

For a given $\epsilon > 0$, there exists a $k \in \mathbb N$ such that for all $n \geq k$, we have $|z - z^\star| < \epsilon$ for all $z \in R_n$

Indeed, for a given $\epsilon > 0$, there exists a $k \in \mathbb N$ such that for all $n \geq k$, we have ${\rm diam \ } R_n < \epsilon$. Now suppose $z$ is contained in $R_n$, where $n$ is any integer $\geq k$. Since $z^\star \in \bigcap_n R_n$, we know that $z^\star$ is also contained in $R_n$ for this value of $n$. Therefore, $|z - z'| < {\rm diam \ } R_n < \epsilon$, as required.
