Union of open sets question

  • #1
117
0
I have to prove that the arbitrary union of open sets (in R) is open.

So this is what I have so far:

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

Any ideas from here?
 

Answers and Replies

  • #2
what is your definition of open set?
 
  • #3
The definition we use is that a set [itex]A\subseteq\mathbb{R} [/itex] is an open set if for each [itex]x\in A[/itex] there exists an [itex]\epsilon>0[/itex] such that [itex](x-\epsilon,x+\epsilon)\subseteq A[/itex].
 
  • #4
note that if [itex]x \in \bigcup_{i \in I}A_i[/itex], then necessarily [itex]x \in A_i[/itex] for some i.

can you continue...?
 
  • #5
Let [itex]\{A_{i\in I}\}[/itex] be a collection of open sets in [itex]\mathbb{R}[/itex]. Let [itex]x\in\bigcup_{i\in I}A_{i}[/itex], then [itex]x\in A_{i}[/itex] for some [itex]i[/itex]. Since each [itex]A_{i}[/itex] is open, there exists an [itex]\epsilon>0 [/itex] such that [itex](x-\epsilon,x+\epsilon)\subseteq A_{i}\subseteq\bigcup_{i\in I}A_{i}[/itex]. Thus, [itex]\bigcup_{i\in I}A_{i}[/itex] is open...

Am I on the right track?
 
  • #6
looks good to me.
 

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