Question

I given a set A. Prove A' = {x : x is an interior point of A} is open. No clue :/

Answer 1

look at you!

Answer 2

lol

Answer 3

i'm pretty sure the start is to let x be an element of A'

Answer 4

It depends on your definition of interior. If your definition of interior is the union of all open sets contained in \(A\) then the interior is by definition open.

Answer 5

The definition of an interior point in this context is there is a neighborhood around x such that x is a subset of a set

Answer 6

sorry such that that neighborhood is subset of a given set

Answer 7

Ok. So here is what I had. Let x be an element of A'. Since x is an interior point of A by definition, there is a neiborhood, called N of x, such that N is subset of A

Answer 8

The goal now is to show there is a neighborhood of x that is subset of A'

Answer 9

Here is what i'm thinking but not sure if it's valid. Given a lemma that proved neighborhoods are open, then every point of N is also an interior point o N. Then there is another neighborhood around x, call M such that M is subset of N

Answer 10

Then I looked at N intersect M

Answer 11

Which appears that M is also subset of A, but I don't know the justification for that

Answer 12

but that didn't help anything since I want N to be subset of N' :/

Answer 13

subset of A' *

Answer 14

did you mean x is an element of A'?

Answer 15

I'm sorry but was x in element of A or A' in your argument?

Answer 16

Suppose \(x \in A'\). Then we have some \(B_\epsilon(x) \subset A\) for \(\epsilon > 0\). Now take some \(\delta < \epsilon\) where \(\delta > 0\) and consider the ball \(B_\delta(x)\), where clearly \(B_\delta(x) \subset B_\epsilon(x)\). Why must every element in the smaller ball be an interior point? Using the triangle inequality you can show that there are balls around every point in the \(\delta\) ball contained in \(B_\epsilon(x)\).

We want to show that \(B_\delta(x) \subset A'\). So, lets take some \(y \in B_\delta(x)\) and show that there is some ball around \(y\) contained in \(A\). In particular we will show there is a ball around \(y\) in \(B_\epsilon(x)\).

So take any \(\gamma < |\epsilon − \delta|\) and consider \(B_\gamma(y)\). Let \(z \in B_\gamma(y)\) and notice the following.
$$\epsilon = \delta + (\epsilon − \delta) > |y − x| + |z − y| \geq |z − x|$$
Therefore \(z \in B_\epsilon(x) \) and therefore \(B_\gamma(y) \subset B_\epsilon(x) \subset A\) and therefore \(y \in A′\).

Answer 17

If any part of my argument is unclear please let me know. I will try to provide a better explanation.

Answer 18

i'm still trying to understand your argument.

Answer 19

Which part of the argument are you having trouble with?

Answer 20

I don't get why the triangle inequality shows \(B_{\delta}(x) \subset A'\).

Answer 21

You need to look carefully at the second part of the argument. We consider an arbitrary point \(y \in B_\delta(x)\) and show that there is a ball around \(y\), \(B_\gamma(y) \subset B_\epsilon(x) \subset A\). The triangle inequality is applied on this line of the proof:
\(\displaystyle\epsilon = \delta + (\epsilon − \delta) > |y − x| + |z − y| \geq |z − x|\)

Answer 22

Since there is a ball \(B_\gamma(y) \subset A\), by the very definition of "interior point". We know that \(y \in A'\).

Answer 23

ok, I'll take a look at the argument again. Thank you for your replies.

Answer 24

Read each line carefully and tell me if there is something you don't understand.