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The Definitions: A family $\{ A_s \}_{s \in S}$ of subsets of a topological space $X$ is locally finite if for every point $x \in X$ there exists a neighbourhood $U$ such that the set $\{ s \in S \mid U \cap A_s \ne \emptyset \}$ is finite. If every point $x \in X$ has a neighbourhood that intersects at most one set of a given family, than we say that the family is discrete. Clearly any discrete family, as well as any finite family, is locally finite. A family of subsets of a topological space is called $\sigma$-locally finite ($\sigma$-discrete) if it can be represented as a countable union of locally finite (discrete) families.

Now I want to show that a locally finite open cover of a metrizable space is not necessarily $\sigma$-discrete.

For this I consider the discrete topology on $\mathbb R$, now $$ \mathbb R \subseteq \bigcup_{X \subseteq \mathbb R} X $$ is an open cover, this cover is just the set $2^{\mathbb R}$. Now because $\{ x \}$ is open in the discrete topology and $|\{x\} \cap U| \le 1$ for every $U\subseteq \mathbb R$, this cover is locally finite. I guess it is not $\sigma$-discrete, cause if $$ 2^{\mathbb R} = \bigcup_{i=1}^{\infty} L_i $$ with $L_i$ being discrete seems impossible for me, but I have no idea how to proof this, maybe derive some kind of contradiction to the uncountabililty of $\mathbb R$... any ideas?

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  • $\begingroup$ The cover is not locally finite, each $x \in \mathbb{R}$ is contained in infinitely many $U \subset \mathbb{R}$. The definition of local finiteness say "each point has a neighbourhood that meets only finitely many sets of the family". $\endgroup$ – Daniel Fischer Jul 10 '13 at 17:31
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The idea of using a discrete space can be salvaged. If $X$ has the discrete topology, we want a point-finite open cover $\mathscr{U}$ of $X$ such that if $\mathscr{U}=\bigcup_{n\in\Bbb N}\mathscr{U}_n$, then there is an $n\in\Bbb N$ such that $\mathscr{U}_n$ is not a pairwise disjoint family.

Let $X$ be an uncountable set with the discrete topology, and let $\mathscr{F}$ be the set of non-empty, finite subsets of $X$. Let $\varphi:X\to\mathscr{F}$ be any bijection. For $y\in X$ let $U_y=\{x\in X:y\in\varphi(x)\}$, and let $\mathscr{U}=\{U_y:y\in X\}$. Then for any $x,y\in X$, $x\in U_y$ iff $y\in\varphi(x)$, so $x$ belongs to exactly $|\varphi(x)|$ many members of $\mathscr{U}$. Since $x$ is isolated and $|\varphi(x)|$ is finite and non-zero, $\mathscr{U}$ is locally finite.

However, for any $y,z\in X$ we have

$$U_y\cap U_z=\big\{x\in X:\{y,z\}\subseteq\varphi(x)\big\}=\big\{\varphi^{-1}[F]:\{y,z\}\subseteq F\subseteq X\text{ and }F\text{ is finite}\big\}\;,$$

so $U_y\cap U_z$ is not just non-empty, but even uncountable. Since $\mathscr{U}$ is uncountable, and no subset of $\mathscr{U}$ with more than one member is pairwise disjoint, $\mathscr{U}$ is not $\sigma$-discrete.

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  • $\begingroup$ why you think there could be a bijection $\phi : X \to \mathscr F$. Guess there are much more finite subsets then elements in $X$, because for every $x \in X$ the subset $\{ x \} \in \mathscr F$, and even much more... $\endgroup$ – StefanH Jul 11 '13 at 14:07
  • $\begingroup$ @Stefan: It’s a standard, basic fact about infinite cardinalities that if $X$ is an infinite set, and $\wp_{\text{fin}}(X)$ is the set of finite subsets of $X$, then $|X|=|\wp_{\text{fin}}(X)|$, meaning that there is a bijection between $X$ and $\wp_{\text{fin}}(X)$. $\endgroup$ – Brian M. Scott Jul 11 '13 at 18:33
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The family you suggest isn't locally finite as Daniel Fischer notes above. However, below I've outlined a proof that, in any case, the family you suggest isn't $\sigma$-discrete.

A discrete family with respect to the discrete topology on $\mathbb{R}$ must be a family of disjoint subsets of $\mathbb{R}$. Let $L_i$ be a family of disjoint subsets of $\mathbb{R}$ for each $i\in \mathbb{N}$. We wish to show that ${\cal P}(\mathbb{R})\neq \bigcup_{i=1}^{\infty} L_i$.

Let $L_i$ consist of the subsets $L_{i}^{\alpha}$ for $\alpha\in S_i$, $S_i$ an index set for each $i\in \mathbb{N}$. Let's assume that there's at least two nonempty $L_{i}^{\alpha}$ for each $i$; we can simply discard the $i$'s for which this isn't the case because we'll end up discarding at most countably many sets and this is irrelevant for cardinality purposes.

Let $X\subseteq \mathbb{R}$ be obtained by picking exactly one element from each $L_{i}^{\alpha}$; we use the axiom of choice to do this. Prove that $X\not\in \bigcup_{i=1}^{\infty} L_i$. We can find uncountably many such $X$ by simply varying the element we pick from some uncountable $L_{i}^{\alpha}$ and picking the elements from the other $L_{i}^{\alpha}$'s according to the axiom of choice. (If all the $L_{i}^{\alpha}$'s are countable, then their union can't be $\mathbb{R}$ so there is a singleton set not in $\bigcup_{i=1}^{\infty} L_i$ and no further argument is necessary.)

I hope this helps!

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