# why is each open set an $F_\sigma$?

In Royden's Real Analysis:

On $\mathbb{R}$, a set which is a countable union of closed sets is called an $F_\sigma$. Thus every countable set is an $F_\sigma$, as is, of course, every closed set. A countable union of sets in Fa is again in $F_\sigma$. Since $$(a, b)= \cup_{n=1}^\infty [a + 1/n, b - 1/n],$$ each open interval is an $F_\sigma$, and hence each open set is an $F_\sigma$.

Each open interval is an $F_\sigma$, but why is each open set an $F_\sigma$? An open set can be uncountable union of open intervals.

If we are considering a general Borel sigma algebra instead of $B(\mathbb R)$, will each open set be an $F_\sigma$?

Thanks!

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With regard to your second question, I think you mean a ‘topology’ instead of a ‘Borel $\sigma$-algebra’. You have to separate topology from measure theory here because your question is purely topological in nature. –  Haskell Curry Jan 29 '13 at 22:29
@HaskellCurry: Is studying $F_\sigma$ and $G_\delta$ a topic of purely topological? –  Ethan Jan 29 '13 at 22:32
Yes. Their definitions are purely topological. –  Haskell Curry Jan 29 '13 at 22:33
Consider the topology on $\mathbb R$ where a set is open iff it is either empty, or if its complement is countable. A closed set is either all of $\mathbb R$, or a countable (possible finite) set. No open set other than $\mathbb R$ is $F_\sigma$. –  Andres Caicedo Jan 29 '13 at 22:34
@HaskellCurry: Are they of any use in purely topological theory? Are countable operations studied in topology? –  Ethan Jan 29 '13 at 22:35

Every open set in $\Bbb R$ can be written as a union of open intervals with rational endpoints, and there are only countably many such open intervals. Each of them is an $F_\sigma$, so every open set is an $F_\sigma$.

More generally, let $\langle X,d\rangle$ be a metric space, and let $U\subseteq X$ be open and non-empty. For $n\in\Bbb N$ let $F_n=\{x\in U:d(x,X\setminus U)\ge 2^{-n}\}$; each $F_n$ is closed, and $U=\bigcup_{n\in\Bbb N}F_n$. Thus, every open set in a metric space is an $F_\sigma$.

Now let $X$ be the space $\omega_1$ with the order topology. The set $U$ of isolated points of $X$ is open and uncountable. $X$ is countably compact, however, so the only closed subsets of $U$ are the finite subsets, and $U$ cannot be an $F_\sigma$.

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If we are considering a general Borel sigma algebra instead of $B(\mathbb R)$, will each open set be an $F_\sigma$? –  Ethan Jan 29 '13 at 22:25
@Ethan: You mean on an arbitrary space? –  Brian M. Scott Jan 29 '13 at 22:26
the sigma algebra generated by any topology over any set –  Ethan Jan 29 '13 at 22:30
@Ethan: I’ve expanded the answer to show that the answer is no in general. –  Brian M. Scott Jan 29 '13 at 22:35
Open sets in a normal space are $F_\sigma$ if and only if the space is perfectly normal –  Martin Jan 29 '13 at 22:35

As a slight expansion of Brian Scott's argument every open set in $\mathbb R$ is the disjoint union of a countable number of open intervals. In particular if $U$ is an open subset of $\mathbb R$ then for each $x \in U$ we can consider the largest interval $x \in I_x \subset U$. Then the set $\mathcal I=\{I_x : x \in U\}$ is a disjoint open cover of $U$ whose union is precisely $U$. We see that it's countable from observing that each interval contains a rational so there's an injection from $\mathcal I$ to the rationals.

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This is the way I learned this way back when. open intervals are connected, so if two intersect their union is an open interval, –  Barbara Osofsky Jan 29 '13 at 22:48
@BarbaraOsofsky, There's also a proof of this fact in Royden not too much further I imagine than where the OP currently is in the book. –  JSchlather Jan 30 '13 at 0:06