I am studying Measure Theory and I am not understanding how I could state that the set $$\{(x,y):x=y, 0\leq x,y\leq 1\}$$
can be written as
$$\bigcap^\infty_{n=1}\left(\bigcup^{2^n-1}_{k=0} \left[\dfrac{k}{2^n},\dfrac{k+1}{2^n}\right]^2\right).$$
Could someone gives me an intuitive thought about this?
Thank you in advance.
Since you want an intuitive idea - does the following picture help?
The red line is the set $A := \{(x,y):x=y, 0\leq x,y\leq 1\}$, while the blue set is $B_n := \bigcup^{2^n-1}_{k=0} \left[\dfrac{k}{2^n},\dfrac{k+1}{2^n}\right]^2$ for $n=0,1,2,3$.
As you can see, the $B_n$ form a chain $B_0 \supseteq B_1 \supseteq B_2 \supseteq B_3 \supseteq B_4 \dots \supseteq A$ which get closer and closer to just beeing $A$ as $n$ increases.
If you want a formal proof for $A = \bigcap_{n=0}^\infty B_n$, you can do that by showing that $A \supseteq B_n$ for all $n$ and for every $(x,y) \notin A$ there exists some $n$ such that $(x,y) \notin B_n$.