Given that $N \cap F^c\neq\emptyset$, when can we say that $N \cap F=\emptyset$?

$N$ and $F$ are sets in a metric space. $F^c$ is the complement of $F$.

A little bit of context: I am trying to prove that $F^c$ is closed. If we can show that $N \cap F=\emptyset$ in general, the proof will be complete.

A thought: $N \cap F^c=\emptyset \Leftrightarrow N \subseteq F$. Therefore, $N \cap F^c\neq\emptyset \Leftrightarrow N \nsubseteq F$. Is this correct? This would mean that $\color{blue} {N \cap F\neq N} $. At most, $N \cap F\subseteq N$.

With the information given, I don't think we can conclude that $N \cap F=\emptyset$. We can go the other way (i.e. if $N \cap F=\emptyset$, then $N \cap F^c\neq\emptyset$), but this doesn't help with the proof.

  • $\begingroup$ Knowing $ N\cap F^c\neq\emptyset$ tells you absolutely nothing about $ N\cap F$. The question is equivalent to the following: "If it's raining, when is $ N\cap F=\emptyset$? $\endgroup$ – guestDiego Jun 5 '16 at 1:06

Let's say $(E,d)$ is your metric space and $N, F, F^c \subseteq E$. Clearly if $p \in E$ we must have that $p \in F$ or $p \in F^c$ but not both. Let $x \in N$. If we suppose $N \cap F = \emptyset$ then clearly we must have that $x \not \in F$. Where must $x$ be then? What does this imply about $N$ and $F^c$?

  • $\begingroup$ This tells us that $N \subseteq F^c$. You are assuming that $N \cap F = \emptyset$ but that is the result we want. $\endgroup$ – Bourbaki Jun 5 '16 at 1:56
  • $\begingroup$ @SpaceTime There's no issue here. If you start with the assumption that $N \subseteq F^c$ you then obtain an if and only if statement. $\endgroup$ – Mnifldz Jun 5 '16 at 2:30

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