Is the Arbitrary Union of Open Sets in R Open?

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Discussion Overview

The discussion revolves around the proof that the arbitrary union of open sets in the real numbers is itself an open set. It focuses on the definitions and properties of open sets in the context of real analysis.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant seeks to prove that the arbitrary union of open sets in \(\mathbb{R}\) is open and initiates the discussion by asking for ideas.
  • Another participant asks for the definition of an open set, prompting clarification on the concept.
  • A definition is provided, stating that a set \(A \subseteq \mathbb{R}\) is open if for each \(x \in A\), there exists an \(\epsilon > 0\) such that \((x - \epsilon, x + \epsilon) \subseteq A\).
  • It is noted that if \(x \in \bigcup_{i \in I} A_i\), then \(x\) must belong to at least one of the sets \(A_i\).
  • A participant continues the proof by stating that since \(x\) is in some \(A_i\), and each \(A_i\) is open, there exists an \(\epsilon > 0\) such that \((x - \epsilon, x + \epsilon) \subseteq A_i\), which implies it is also contained in the union.
  • One participant expresses agreement with the proof approach presented.

Areas of Agreement / Disagreement

Participants generally agree on the approach to proving that the arbitrary union of open sets is open, though the discussion is still in progress and no final consensus has been reached.

Contextual Notes

The discussion does not address potential limitations or assumptions in the proof, nor does it explore any counterexamples or alternative definitions of open sets.

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I have to prove that the arbitrary union of open sets (in R) is open.

So this is what I have so far:

Let \{A_{i\in I}\} be a collection of open sets in \mathbb{R}. I want to show that \bigcup_{i\in I}A_{i} is also open...

Any ideas from here?
 
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what is your definition of open set?
 
The definition we use is that a set A\subseteq\mathbb{R} is an open set if for each x\in A there exists an \epsilon>0 such that (x-\epsilon,x+\epsilon)\subseteq A.
 
note that if x \in \bigcup_{i \in I}A_i, then necessarily x \in A_i for some i.

can you continue...?
 
Let \{A_{i\in I}\} be a collection of open sets in \mathbb{R}. Let x\in\bigcup_{i\in I}A_{i}, then x\in A_{i} for some i. Since each A_{i} is open, there exists an \epsilon>0 such that (x-\epsilon,x+\epsilon)\subseteq A_{i}\subseteq\bigcup_{i\in I}A_{i}. Thus, \bigcup_{i\in I}A_{i} is open...

Am I on the right track?
 
looks good to me.
 

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