# Derived Set Proof

• Oct 16th 2009, 05:44 PM
Majialin
Derived Set Proof
Hello, I am currently studying for my analysis midterm, and I can't seem to figure out how to solve the following:

Let S,TRn. Show that (SUT)' = S'UT'.

Where S' and T' represent derived sets.

Any help with this problem would be greatly appreciated!
• Oct 16th 2009, 06:47 PM
rn443
Quote:

Originally Posted by Majialin
Hello, I am currently studying for my analysis midterm, and I can't seem to figure out how to solve the following:

Let S,TRn. Show that (SUT)' = S'UT'.

Where S' and T' represent derived sets.

Any help with this problem would be greatly appreciated!

x is an accumulation point of (SUT)' iff there exists a sequence of distinct points a_n of SUT with |x - a_n| < 1/n. This is true iff either infinitely many of the points belong to S or infinitely many belong to T (or both), which is true iff either x belongs to S' or x belongs to T'. Hence (SUT)' = S'UT'.
• Oct 16th 2009, 08:22 PM
Plato
You have posted this question twice. That is strictly against forum rules.
But I will respond this way. There is only one way that the implication is problematic.
That is to prove that $\displaystyle \left( {S \cup T} \right)^\prime \subseteq S' \cup T'$.
If $\displaystyle x$ is limit point of $\displaystyle S\cup T$ and $\displaystyle x$ is limit point of $\displaystyle S$ then we are done.

But if and $\displaystyle x$ is not limit point of $\displaystyle S$ you must prove that $\displaystyle x$ must a limit point of $\displaystyle t$.