Question

Hi, i have a real analysis question:
1. Using the definition , show that the following sets are open:
E = {xeR:absolute value(2x-3)<5}

E = {xeR:1 11 years ago

Answer 1

Well, every open interval in R is open. Your two sets of definition give two open intervals, so that's enough in a non-proofy way. To start ya off on a proof, if you consider an interval (a, b), let p be in (a, b) and let \(\epsilon\) = min {p-a, b-p}. Try and see if you can do some manipulation with that.

Answer 2

The hardest problem is when we have to prove a trivial stuff. :(

Answer 3

That's when you're hoping you can use a trivial theorem of some sort, lol.

Answer 4

Im really lost

Answer 5

Well, what's the criteria needed for your set to be open?

Answer 6

am i supposed to do x=4 andx=-1?

Answer 7

(-1,4) would be your interval for the first set, yes.

Answer 8

I just putting it in the arbitrary case where a and b are neither negative infinity or infinity.

Answer 9

But for the second one given that we have 2 values already given I<m a little confused

Answer 10

You would have to do it for both intervals in the 2nd case. The first interval is (-1,1) and the 2nd interval is (3,5)

Answer 11

But given that (-1,1) is not betwwen 1 and 5 is it still open ?

Answer 12

Are these not separate problems?

Answer 13

I meant 3 sorry

Answer 14

Well, the union of open subsets is also open.

Answer 15

Pardon me for butting in, but is this not two questions? The solution interval for the first set is (-1, 4) and for the second set is (-1, 5). Because both sets are open intervals, then they are open sets.

Answer 16

@ospreytriple I misread exactly what you did. hehehe, the second one is \(1<|x-2|<3\)

Answer 17

OK. Sorry I'm outta here.

Answer 18

Yeah, I was interpretting this as two separate questions. Both sets are open intervals and any open interval in R is open. It's kind of trivial. The union of open subsets of R is also open. Clearly both of these sets would fit the criteria necessary. I'll wait for your word though, lol.

Answer 19

Ohh ok so therefore the answer would be yes because (-1,1)U(3,5) because.. Just for my clarification lol (-1,1) the 1 is in the set and (3,5) is open because 3 belongs to the set

Answer 20

Wel, in the end, the answer is that both sets are open because they give open intervals and any open interval in R is open. If you had to do more work, you would have to show that there exists \(\epsilon >0\) such that the epsilon neighborhood about all points p in your interval are a subset of E. That's where I was then throwing in all the weird epsilon min stuff, lol.

Answer 21

I see ..
this was really helpful thanks!!

Answer 22

But was what i wrote plausible though ?

Answer 23

Well, you said 1 was in the set (-1,1) and 3 was in the set (3,5), which is not true. Before we jump to something more proofy, let's make sure we understand what's going on.

So we want to show that no matter what neighborhood we take about an arbitrary point in our set, our neighborhood will still contain a point in that set.




No matter what point you choose in (a,b), I will have a neighborhood about that point that will also have points within (a,b). Now, I may also have exterior points, but that doesn't matter. What's nice is that we're dealing with R. If we were dealing with something like the set of rationals, this would not be true since we could find neighborhoods that contain irrational numbers not in the set. Does that make sense so far?