Proving Open Sets: Definition and Example in R3 | Homework Solution

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Homework Statement

Define the notion of an open set in Rn.
Prove that the set {(x⁴)(y⁴)(z) | x > 0; y > 0; z < 0 } is open in R3.

Homework Equations

Definition of an open set

B_r(p) = { x is an element in Rⁿ |x - p| < r} c U

The Attempt at a Solution

Well we first pick a point p in U, and we want to find and r > 0 s.t. B_r(p) is in U.
Choose r = min(|x⁴||y⁴||z|)

Then q= (x`, y`, z`) element of B_r(p)

(x` - x<sup>4</sup>)<sup>2</sup> + (y` - y⁴)² + (z` - z)² < r²

And from here I have to show x` > 0, y` > 0, z` < 0
x and y will follow the same way procedurally; from the inequality above, we know

(x` - x<sup>4</sup>)<sup>2</sup> < r<sup>2</sup>
|x` - x⁴| < r

Here is where I get stuck; with exponents I am finding this more difficult to rearrange, am I supposed to use this inequality;?

x² + y² >= 2xy, by comparing both x and y to r?

Any help or insight would be great thanks, proving open sets have been giving me a lot of difficulty.

 

[foxjwill]

First of all, a tip on writing proofs: Make sure, in general, to define all your symbols and variables. Although the reader might be able to figure out what your notation means based on previous knowledge, most would rather spend the majority of their time understanding your proof.

For example, your definition for an open set uses no standard symbols, and it took me a minute or so to figure out that you were trying (i think) to define an open set in $$\mathbb{R}^n$$ as follows: A subset $$U$$ of $$\mathbb{R}^n$$ is called open if for every point p in U there's an r>0 such that the set $$B_r(p)=\{x\in\mathbb{R}^n : |x-p|<r\}\subseteq U.$$

Also, I'm assuming $$(a)(b)(c)$$ is the ordered triplet $$(a,b,c)$$

As to the proof, the easiest way to go about this is to realize that the set $$\{(x^4,y^4,z):x>0,y>0,z<0\}$$ is the same as the set $$\{(x,y,z):x>0,y>0,z<0\}$$. Once you do that, just continue the proof in the same way as before and you should be fine.

 

[Iconate]

Hey thanks, sorry for my syntax, first time on this forum, getting used to latex.

I am not too sure how they are equivalent; but, does this follow correctly;

I had
(x` - x⁴)² < r²

and instead of saying;
|x` - x⁴| < r

Can I say; r² $$\leq$$ x⁴
thus;

|x` - x⁴| < x⁴

Adding x⁴ to each;

0 < x` < 2x⁴

And y` will follow the same way, however is that correct in saying that x` and y` are open?
(still have to show z` is open)

 

[foxjwill]

Hey thanks, sorry for my syntax, first time on this forum, getting used to latex.

I am not too sure how they are equivalent; but, does this follow correctly;

Suppose $$(x^4,y^4,z)\in \{(x^4,y^4,z):x>0,y>0,z<0\}$$. Then x^4>0, y^4>0, and z<0, so $$(x^4,y^4,z)\in\{(x,y,z):x>0,y>0,z<0\}$$.

Conversely, if $$(x,y,z)\in\{(x,y,z):x>0,y>0,z<0\}$$, then $$(x,y,z)=\left((\sqrt[4]{x})^4,(\sqrt[4]{x})^4,z\right)\in\{(x^4,y^4,z):x>0,y>0,z<0\}$$. Thus, the two sets are equivalent.

Can I say; r² $$\leq$$ x⁴
thus;

|x` - x⁴| < x⁴

Adding x⁴ to each;

0 < x` < 2x⁴

And y` will follow the same way, however is that correct in saying that x` and y` are open?
(still have to show z` is open)

If you meant $$r\leq x^4$$, then your reasoning is correct (assuming by $$r=\min(|x^4||y^4||z^4|)$$ you meant $$r=\min\{|x^4|,|y^4|,|z^4|\}$$). However, x' is not open, nor is (x',y',z)--it is a point contained in the set $$B_r(p)$$ which is contained in some open set U.

Also, a semicolon is not the same as a colon.

 

[Iconate]

If you meant $$r\leq x^4$$, then your reasoning is correct (assuming by $$r=\min(|x^4||y^4||z^4|)$$ you meant $$r=\min\{|x^4|,|y^4|,|z^4|\}$$). However, x' is not open, nor is (x',y',z)--it is a point contained in the set $$B_r(p)$$ which is contained in some open set U.

Also, a semicolon is not the same as a colon.

Yes, sorry I did mean $$r\leq x^4$$
Why isn't x` open when I showed that x` > 0 (forgetting the other two points y`, and z` at the moment)

 

[foxjwill]

Why isn't x` open when I showed that x` > 0 (forgetting the other two points y`, and z` at the moment)

Recall that a subset U of $$\mathbb{R}^n$$ is defined to be open if for every point in U there's an r>0 such that $$B_r(p)=\{x\in\mathbb{R}^n : |x-p|<r\}\subseteq U$$.

In other words, the requirements for some object U to be open in $$\mathbb{R}^3$$ are that

1. U is a set,
2. U is a subset of $$\mathbb{R}^3$$, and
3. for every point in U there's an r>0 such that $$B_r(p)=\{x\in\mathbb{R}^3 : |x-p|<r\}\subseteq U.$$

The object x' doesn't satisfy any of these.

Now recall that the elements of $$\mathbb{R}^3$$ are called points. Since x' is not an element of $$\mathbb{R}^3$$, it is not a point (and neither are y' or z'). However, the ordered triple (x',y',z') is an element of $$\mathbb{R}^3$$ and is therefore a point. But is it an open set? No, it is not, because it again doesn't satisfy any of the requirements above.

Finally, consider the set $$\{(x^4,y^4,z):x>0,y>0,z<0\}$$. Clearly, it satisfies the first two requirements. Thus, what you've been trying to show in your proof is that it satisfies the third and is therefore open. But to reiterate, the elements (i.e. the points) in that set are not open.