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Separability: Does it require that the finite union of e.g. open intervals are non-disjoint?

E.g. when proving that $L_1([0,1])$ is separable.

Intuitively the open intervals themselves are dense, since for each pair of rationals, there's always more between them.

But what about finite unions of them? Intuitively if the unions were non-disjoint, then the resulting set would be dense. However, when constructing dense sets, I've seen the opposite, that they may not explicitly require non-disjointness.

But how are finite disjoint unions of open intervals with rational endpoints dense?

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  • $\begingroup$ Dense in what? What is your ambient space? $\endgroup$ – Landon Carter Feb 11 at 11:22
  • $\begingroup$ @LandonCarter $L_1([0,1])$ for example. The construction of density for this space I believe depends on the construction of such union, but I don't understand how it can make it dense, if the union contains disjoint sets. Or perhaps one requires that the sets are joined? $\endgroup$ – mavavilj Feb 11 at 11:25
  • $\begingroup$ @mavavilj What does "construction of density for this space" mean? $\endgroup$ – 5xum Feb 11 at 11:27
  • $\begingroup$ @5xum That one constructs an argument for why $L_1([0,1])$ contains a countable dense subset. Or i.e. that $L_1([0,1])$ is separable. $\endgroup$ – mavavilj Feb 11 at 11:28
  • $\begingroup$ @jjagmath I don't know how you know that. The question is about proving that $L_1([0,1])$ is separable (a topological property) and is tagged "general-topology". $\endgroup$ – 5xum Feb 11 at 11:32
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This is too long for a comment, so...

Your answer cannot be answered because the following is not clear:

What do you mean by "dense" here?

To explain:

Usually, "dense" is a property of a single set with regard to a topological space. For example, $(0,1)$ is dense in $[0,1]$, but is not dense in $\mathbb R$. In particular, denseness is defined as follows:

If $X$ is a topological space and $S\subseteq X$ is a subset of $X$, then $S$ is dense if and only if the closure of $S$ is equal to $X$, i.e. $\overline S=X$.

Using the standard meaning of "dense", a finite union of open intervals, in general, is not dense in $\mathbb R$. For example, $(0,1)\cup(2,3)$ is a finite union of open intervals, but it is not dense. I understand this is not what you wanted to ask, but unfortunately, that is what you asked, and a better answer would require you to explain your question better. I advise you to edit your question so it asks what you really want to know.

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