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When can I say that $A=\text{cover of A}$?

The definition of cover seems that a cover of $A$ is some $\bigcup_j C_j$ of "covers" $C_j$, s.t.

$$A \subset \bigcup_j C_j$$

I think in some cases it's possible to say directly that $\subset$ is $=$. But in cases, when it's not "trivial" or if one wants to be sure, then what to do?


Perhaps:

"=>" If I take $x \in A$, then I can find it from $\bigcup_j C_j$.

"<=" If I take $x \in \bigcup_j C_j$, then I can find it from $A$.

However, how does one argue that an infinite union actually contains the required element?

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  • $\begingroup$ When you need equality and it's not trivial you have to work with the details of that particular cover to prove what you need. There no generic way to argue. $\endgroup$ Sep 8, 2019 at 10:15
  • $\begingroup$ @EthanBolker I'd assume that there'd be cases for $\mathbb{R}$ which generalize to higher dimensions or subsets of $\mathbb{R}$. At least when working in $\mathbb{R}^n$. $\endgroup$
    – mavavilj
    Sep 8, 2019 at 10:17
  • $\begingroup$ I don't understand your comment. Please edit your post to include a particular cover or example that led you to ask this general question. Then perhaps we can help. Further back and forth in comments won't be much use. $\endgroup$ Sep 8, 2019 at 11:55

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$\{C_j\}_{j\in \mathbf{N}}$ covers $A$ if $A \subseteq \bigcup_{j \in \mathbb{N}} C_j$

To show equality you have to prove that $\bigcup_{j \in \mathbb{N}} C_j \subseteq A$ and this happens if $\forall \ x \in \bigcup_{j \in \mathbb{N}} C_j\Rightarrow x \in A $

For example, consider $A:=[0,1]$, then $[n,n+1], \ n\in \mathbb{Z}$ is obviously a cover of $A$ but they are not the same set.

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  • $\begingroup$ and $[\frac{1}{n},1]$ for $n\in\mathbb{N}$ is a cover of $A$ which equals to $A$. $\endgroup$
    – Yanko
    Sep 8, 2019 at 15:43
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A cover of $A$ is a family of sets whose union includes $A$. The simplest example of a cover of $A$ is the family $\{A\}$ with only one set in it.

So I say: YES $\{A\}$ is a cover of $A$

but

NO $A$ is not a cover of $A$.

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