# Why the metric space is not compact?

Let $(X,M, \mu)$ a finite measure space and define a metric space $M'$ as follows: for $A,B \in M$, define: $d(A,B) = \mu(A\triangle B)$.

The space $M'$ is defined as all sets in $M$ where sets $A$ and $B$ are identified if $\mu(A\triangle B) = 0$.

I yet proved that $(M', d)$ is complete metric space.

Question: Show that the space above is not compact when $X = [0, 1]$, $M$ is the family of Borel sets on $[0, 1]$, and $\mu$ is Lebesgue measure.

I know that for any metric space $(X,d)$, $(X,d)$ is compact iff is complete and totally bounded. By the above result the metric space of question is complete then i should show that is not totally bounded. But i have not idea how proceed. This is the best way for to prove? I know also that i could to find a sequence that have not subsequence that converge in $M$, but i have not found. Any tip for the sequence?

• A metric space is compact if and only if it is sequentially compact. Try to find a sequence of sets with $d(A_m, A_n) = \frac{1}{2}$ for all $m \neq n$. – Daniel Fischer Feb 10 '17 at 12:36
• This is possible? – user402085 Feb 11 '17 at 3:28
• Yes. To get you started, take $A_1 = \bigl[0,\frac{1}{2}\bigr)$ and $A_2 = \bigl[0,\frac{1}{4}\bigr) \cup \bigl[\frac{1}{2},\frac{3}{4}\bigr)$. – Daniel Fischer Feb 11 '17 at 11:08
• I see now. thanks – user402085 Feb 11 '17 at 12:02