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A subset of a topological space X is a $F_\sigma$ if and only if its a countable union of open sets.

Show that the $F_\sigma$ can be written as the union of an increasing sequence of closed sets.

I know that if a set is countable then it will be arranged in a sequence form so $F_\sigma$ can be in a sequence form but what on the union how do we show that?

  1. let X be a metrizable space whose topology is generated by a metric p. how that 2p defined by 2p(x,y) = 2.p(x,y). For this question i know a metrizable space induce a metric but i dont und erstand the metrizable space itself? and i dont know how to go about it
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2 Answers 2

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Here is an elaboration on Henno Brandsma's answer. But first of all we say a subset of a topological space is $F_\sigma$ if it can be written as a countable union of closed sets. Anyway suppose $X=\bigcup_{n=1}^\infty F_n$. Define $E_n=\bigcup_{k=1}^n F_k$. Then $X=\bigcup_{n=1}^\infty E_n$ and the sequence $(E_n)_{n=1}^\infty$ is an increasing sequence. And in any event each $E_n$ is closed (or open if you go with your definition of $F_\sigma)$.

Next, we say a topological space $(X,\tau)$ is metrizable if there exists a metric $\rho:X\times X\to[0,\infty)$ such that the topology induced by the metric $\rho$ coincides with the topology of $X$, i.e. $\tau$. So you can consider $\tau$ to be the topology induced by the metric $p$ and $\tau^\prime$ to be the topology induced by the metric $2p$ in your question. Now you need to show that $\tau$ is the same as $\tau^\prime$. For that consider $$\mathcal{B}=\{B_p(x,r):x\in X,r>0\}$$ and $$\mathcal{B^\prime}=\{B_{2p}(x,2r):x\in X,r>0\}$$ which are bases for $\tau$ and $\tau^\prime$ respectively. Now observe that $B_p(x,r)=B_{2p}(x,2r)$ because $$a\in B_p(x,r)\iff p(x,a)<r\iff 2p(x,a)<2r\iff a\in B_{2p}(x,2r).$$ Hence the two topologies are the same.

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As to the first point

$$F_1 \cup F_2 \cup F_3 \ldots = F_1 \cup (F_1 \cup F_2) \cup (F_1 \cup F_2 \cup F_3) \ldots $$

  1. Show that the topoogy induced by $d$ is the same as the one by $2d$: $B_d(x,r) = B_{2d}(x,2r)$ for all $x \in X, r>0$
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  • $\begingroup$ please for the first point can you give an example to make it easier?? and the 2. i dont understand at all, please any thing to show further? $\endgroup$
    – esther
    Commented Sep 14, 2017 at 5:48
  • $\begingroup$ The first is the essential idea of the proof. As to 2. what is the definition of the topology from a metric? Cannot you see that $d(x,y) < r$ iff $2d(x,y) < 2r$? Any $d$ ball is a $2d$ ball and vice versa. $\endgroup$ Commented Sep 14, 2017 at 6:07

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