Question

Hi, i have a real analysis question: Using the definition , show that the following sets are closed
E = {xeR: |5x−3|=|2x+4|}

and
E = {xeR:1≤x^2≤3}
Thanks!

Answer 1

Couple claims can be made about this just like in the previous problem. Just like every open interval in R is open, every closed interval in R is closed. Another thing you can add is that if the compliment of E is open, then E is closed. For example, the compliment of [1,3] is \((-\infty,1)\cup(3,\infty)\). Both of those intervals are open, their union is open, thus [1,3] is closed. So that might be the idea to try and use, the compliment one.

Answer 2

first set is finite so i guess we may mimic "finite sets are closed" proof

Answer 3

another method
ok you need to show for E3= E1UE2 , E3 is closed
and intersection of any family of indexed set is also closed

Answer 4

But do i find the intervals in the first one to solve it even though its a finite set ?

Answer 5

Or do i bring the equation onto one side ?

Answer 6

From any notes I have from my class, it just seems like the easiest way would be to show that the compliment is open. In my class, we didnt really do any proofs of sets being closed because by then we were just using theorems that made it trivial. It was just something we were allowed to use in our proofs, closed subsets of R are closed. So thats the main reason why Im quick to go to the compliment idea. The set need not be finite, though. If you're willing to go with the compliment being open idea, it works out very similar to the case where we were finite.

Answer 7

So I find the intervals then the compliment?

Answer 8

Yes. And the compliment works like the example I gave. When you work with an interval \((-\infty, a)\), let epsilon = a-p. When you work with an interval \((b, \infty)\), let epsilon = p-b

Answer 9

but given that both equations are = can i bring the equation on 1 side ?

Answer 10

Oh, for the first set?

Answer 11

Yes can I write 3x-7 = 0 ? Or do i need to find the intervals of both sides ?

Answer 12

For two absolute values being equal, you can choose one of the two to not use absolute value and then use the normal rules. So given we |5x - 3| = |2x + 4|, we can treat this as if it were simply |5x -3| = 2x + 4, which means we would have the two equations 5x -3 = 2x + 4 and 5x-3 = -2x - 4. That'll give us two values that we will want to use.

Answer 13

Ohh I see

Answer 14

I can justify that in this way: \(|x| = \sqrt{x^{2}}\). This means we have
\(\sqrt{(5x-3)^{2}}\ = \sqrt{(2x+4)^{2}}\)
\(\implies (5x-3)^{2} = (2x+4)^{2}\)
\(\implies (5x-3) = \pm (2x+4)\)

Answer 15

I see ..because of the absolute values