I would like to know the meaning of the following:
I need to prove that $\bigcup_{n=1}^{\infty}{A_n}=A$ is true, where A is a set of positive real numbers, and $A_n = [x\space\epsilon\space A: x\geq 1/n]$, where $n\space\epsilon\space\Bbb{N}$. I have two questions. Firstly, does this mean that $A_n$ can contain the same number multiple times? For example, could $A_4=[1/3,1/3]$? Secondly, and most importantly, what does the left hand side of the equality mean? I get the impression that it's the union of all the $A_n$ sets for every natural numbers, but I can't tell.
Thanks in advance.

  • $\begingroup$ A set never contains an element more than once. $A_4=\{x\in A| x\ge 1/4\}$ This has nothing particular to do with $[1/3,1/3]=\{1/3\}$ $\endgroup$ – saulspatz Apr 24 '18 at 15:58
  • $\begingroup$ Where is the screenshot you speak of? $\endgroup$ – JuliusL33t Apr 24 '18 at 16:06
  • $\begingroup$ Someone editted the post, the screenshot contained the equation that is shown at the top. I changed the post. Also thanks for the clarification on the elements of a set. $\endgroup$ – Quaere Verum Apr 24 '18 at 16:09
  • $\begingroup$ Also, I think I found the solution to the problem altogether, so thanks for the clarification :) I think the proof goes along the lines of the following: since 1/(n+1)<1/n, if an element belongs to A_n, it also belongs to A_n+1. Therefore, the union of all the A_n subsets is equal to A_infinity, which contains all the positive numbers in A. Since A contains only positive numbers, it equals A. Does that sound about right? $\endgroup$ – Quaere Verum Apr 24 '18 at 16:26
  • $\begingroup$ That's the right idea, but you should state it more precisely. To prove that two sets $X$ and $Y$ are equal, you need to show that $X \subseteq Y$ and $Y \subseteq X$. In your case, $X = \bigcup_{n=1}^{\infty}A_n$ and $Y = A$. Proving that $X \subseteq Y$ is easy: if $x \in X$ then $x \in A_n$ for some $n$, hence $x \in A$ by definition of $A_n$. Proving that $Y \subseteq X$ isn't much harder: if $x \in Y = A$, then $x$ is a positive number. Hence there is some $n$ such that $0 < 1/n < x$, and therefore $x$ is in the corresponding $A_n$, hence in the union. $\endgroup$ – Bungo Apr 24 '18 at 16:43

Going off @Bungo's comment:

I usually try to avoid using "no so clearly defined" concepts in my proofs, like $A_{\infty}$. To me it just looks like a change in notation, i.e. you're putting

$$A_{\infty} := \bigcup_{n=1}^{\infty}{A_n}$$

which is clearly true since it's a definition. Now the question is whether $A_{\infty} = \mathbb{R}_+$.

Instead we could just prove the two inclusions $\bigcup_{n=1}^{\infty}{A_n} \subset \mathbb{R}_+ $ and $\mathbb{R}_+ \subset \bigcup_{n=1}^{\infty}{A_n} $

If $ x \in \bigcup_{n=1}^{\infty}{A_n}$ then for some $n$, $x \in A_n$. But each $A_n$ contains only positive real numbers, so $x$ must be one too, so $x \in \mathbb{R}_+$

And if $x \in \mathbb{R}_+$, then there is some $n_0$ s.t. $x \geq \frac{1}{n_0}$, so $x \in A_{n_0}$ by definition, and so $ x \in \bigcup_{n=1}^{\infty}{A_n}$.

So the two sets are equal, since they contain the same elements.

  • $\begingroup$ I think the distinction between "a" set of positive real numbers and "the" set of positive real numbers is important here. I'm not sure if it matters for the proof, but I thought I should at least mention that A in my question is an arbitrary set of positive real numbers. $\endgroup$ – Quaere Verum Apr 24 '18 at 20:38
  • $\begingroup$ But the equality isn't true for any arbitrary set of positive real numbers! Put $A = (0,1)$. Then $2\in A_1$, since $2 > 1$, so it's in the union, but $2 \not \in A$. Maybe you mean something else? $\endgroup$ – JuliusL33t Apr 24 '18 at 21:58
  • $\begingroup$ That's not true, the set $A_1 = [x \space\epsilon\space A : x \geq 1]$ which would then only contain one. Every $A_n$ is a subset of A. $\endgroup$ – Quaere Verum Apr 25 '18 at 8:07

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