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I'm trying to prove $(A \cup B)' = A' \cup B'$ where $S'$ denotes the derived set of some subset $S$ of a topological space $X$. A derived set $S'$ of a set $S$ is the set of $x \in X$ such that $x$ is in the closure of $S-\{x\}$. I'm having troubling showing that $(A \cup B)' \subseteq A' \cup B'$.

What I've done so far: Suppose $x \in (A \cup B)'$. Then for any neighborhood $U$ of $x$ it follows that $U$ intersects $(A \cup B)-\{x\}$ so $U$ intersects $A-\{x\}$ or $B-\{x\}$. But I don't see why we couldn't have one neighborhood of $x$, $U_{1}$, which intersects $A-\{x\}$ but not $B-\{x\}$ and another neighborhood $U_{2}$ which does the opposite. In this case $x$ would be in neither $A'$ nor $B'$. We can say that $U_{1} \cap U_{2}$ is not only $\{x\}$ because then $\{x\}$ would be open contradicting $x \in (A \cup B)'$.

goatman2743
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  • @Daniel: I disagree, this is not about closure, it's about derived sets. The questions are fundamentally different anyway. – Asaf Karagila Sep 19 '13 at 19:39
  • If you assume that some neighbourhood $U$ of $x$ does not intersect $A\setminus{ x}$, then any smaller neighbourhood $V$ has the same property; but since $x\in (A\cup B)'$, it follows that any $V\subset U$ intersects $B\setminus{ x}$, and hence $x\in B'$. – Etienne Sep 19 '13 at 19:50

1 Answers1

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Ah, just noticed this is exactly Etienne's comment.

If all neighborhoods of $x$ intersect both $A\setminus\{x\}$ and $B\setminus\{x\}$, then we are done. Thus, assume to the contrary that a neighborhood $U$ intersects only $A\setminus\{x\}$. We will prove that $x\in A'$.

Suppose $V$ is any neighborhood of $x$ and note that $U\cap V$ is a neighborhood of $x$. Since $U\cap V\cap(B\setminus\{x\})\subseteq U\cap(B\setminus\{x\})=\emptyset$, it follows that $U\cap V$ intersects $A\setminus\{x\}$. Therefore, $V$ intersects $A\setminus\{x\}$ as desired.

Karl Kroningfeld
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