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?


"=>" 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?

  • $\begingroup$ 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

2 Answers 2


$\{C_j\}_{j\in \mathbf{N}}$ 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.

  • $\begingroup$ 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

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 $A$


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.