Thread: Proving interval is closed in R

1. Proving interval is closed in R

The approach I am supposed to use is to show $\displaystyle \exists p\in [0,\infty)$ such that any open ball $\displaystyle B_r(p)$ centered at $\displaystyle p$ with $\displaystyle r>0$ is not in $\displaystyle [0,\infty)$. The space is in R, so the ball is in R as well. I can express the ball as $\displaystyle B_r(p)=(p-r, p+r)$. The $\displaystyle p$ we are looking for should be $\displaystyle 0$, so the ball is $\displaystyle B_r(0)=(-r,r)$. Clearly any $\displaystyle r>0$ will give an interval with points outside of $\displaystyle [0,\infty)$. I am very bad with proofs, but I don't believe I can just say $\displaystyle (-r,r)$ is not in $\displaystyle [0,\infty)$ since $\displaystyle -r<0$ for $\displaystyle r>0$ and therefore there are points in the ball not in $\displaystyle [0,\infty)$ and be done. How can I finish this?

2. Re: Proving interval is closed in R Originally Posted by adkinsjr The approach I am supposed to use is to show $\displaystyle \exists p\in [0,\infty)$ such that any open ball $\displaystyle B_r(p)$ centered at $\displaystyle p$ with $\displaystyle r>0$ is not in $\displaystyle [0,\infty)$. The space is in R, so the ball is in R as well. I can express the ball as $\displaystyle B_r(p)=(p-r, p+r)$. The $\displaystyle p$ we are looking for should be $\displaystyle 0$, so the ball is $\displaystyle B_r(0)=(-r,r)$. Clearly any $\displaystyle r>0$ will give an interval with points outside of $\displaystyle [0,\infty)$. I am very bad with proofs, but I don't believe I can just say $\displaystyle (-r,r)$ is not in $\displaystyle [0,\infty)$ since $\displaystyle -r<0$ for $\displaystyle r>0$ and therefore there are points in the ball not in $\displaystyle [0,\infty)$ and be done. How can I finish this?
The above does not make any sense. Please state what it is that you are to prove.
From the title is seems to be: Prove an interval is closed. But of course that is not true unless your text has a different definition for interval.
On the other hand. the interval $[0,\infty)$ is closed. Is that what you are to prove?

3. Re: Proving interval is closed in R

Yes, to be clear the goal is to prove $\displaystyle [0,\infty)$ closed.

4. Re: Proving interval is closed in R Originally Posted by adkinsjr Yes, to be clear the goal is to prove $\displaystyle [0,\infty)$ closed.
See how easy that was. Now the standard way to prove a set is closed is by showing its complement is open.
Why is that? Well, because by definition a closed set contains all of its limit points.
If $t\in X^c$ and $\exists \mathcal{O}$, open set, such that $t\in \mathcal{O}~\&~\mathcal{O}\cap X=\emptyset$ then $t$ is not a limit point of $X$
How can you show that $(-\infty,0)$ is open?

5. Re: Proving interval is closed in R Originally Posted by Plato See how easy that was. Now the standard way to prove a set is closed is by showing its complement is open.
Why is that? Well, because by definition a closed set contains all of its limit points.
If $t\in X^c$ and $\exists \mathcal{O}$, open set, such that $t\in \mathcal{O}~\&~\mathcal{O}\cap X=\emptyset$ then $t$ is not a limit point of $X$
How can you show that $(-\infty,0)$ is open?
Thanks for your response.

My approach originally was to use the open complement argument. I asked my professor via e-mail if I should show $\displaystyle (-\infty,0)$ is open and he didn't affirm it in simple terms. He just said to find some t∈[0,∞) such that any open ball Br(t) centered at t with r>0 is not in [0,∞), which I think is equivalent to your explanation.

Definition of openess: For any $\displaystyle t\in (-\infty,0)$ there is an open ball $\displaystyle B_r(t)=x: |x-t|<r$ or just $\displaystyle (t-r , t+r)$ centered at $\displaystyle t$ with r>0 such that the ball is contained in $\displaystyle (-\infty,0)$.

So if $\displaystyle x\in (t-r,t+r)$ then $\displaystyle x\in (-\infty,0)$ needs to be shown. I don't know what trick to use to show that.

6. Re: Proving interval is closed in R Originally Posted by adkinsjr Thanks for your response.

My approach originally was to use the open complement argument. I asked my professor via e-mail if I should show $\displaystyle (-\infty,0)$ is open and he didn't affirm it in simple terms. He just said to find some t∈[0,∞) such that any open ball Br(t) centered at t with r>0 is not in [0,∞), which I think is equivalent to your explanation.

Definition of openess: For any $\displaystyle t\in (-\infty,0)$ there is an open ball $\displaystyle B_r(t)=x: |x-t|<r$ or just $\displaystyle (t-r , t+r)$ centered at $\displaystyle t$ with r>0 such that the ball is contained in $\displaystyle (-\infty,0)$.

So if $\displaystyle x\in (t-r,t+r)$ then $\displaystyle x\in (-\infty,0)$ needs to be shown. I don't know what trick to use to show that.
O.K. Suppose that $t\in(-\infty,0)$ then $|t|>0$ Why is that true?
Define $\delta=\frac{|t|}{2}$ Why is $\delta>0$?
Can you argue that $(t-\delta,t+\delta)\cap(0,\infty)=\emptyset~?$
Now can you explain why that means $t$ is an interior point of $(-\infty,0)~?$
Thus can you explain why that proves that $t$ is not a limit point of $[0,\infty)~?$

7. Re: Proving interval is closed in R Originally Posted by Plato O.K. Suppose that $t\in(-\infty,0)$ then $|t|>0$ Why is that true?
Absolute values are always greater than 0. Originally Posted by Plato Define $\delta=\frac{|t|}{2}$ Why is $\delta>0$?
Because $\displaystyle |t|>0$ Originally Posted by Plato Can you argue that $(t-\delta,t+\delta)\cap(0,\infty)=\emptyset~?$
If we use $\displaystyle \delta =| t|/2$ in the interval? $\displaystyle \left(t-\frac{t}{2}, t+\frac{t}{2}\right)=\left(\frac{t}{2},\frac{3t}{2 }\right)$.

I'm not understanding this, and why are we looking for the intersection with $\displaystyle (0,\infty)$?

EDIT:

I think I see what you meant, the intersection being empty just means there are no points $\displaystyle t$ in the ball $\displaystyle (t-\delta,t+\delta)$ that are also in the interval $\displaystyle [0,\infty)$. So $\displaystyle t$ must be in the complement $\displaystyle (-\infty,0)$. I still do not understand how to show it's empty using your construction.

8. Re: Proving interval is closed in R Originally Posted by adkinsjr Because [math]|t|>0[/math

If we use $\displaystyle \delta =| t|/2$ in the interval? $\left(t-\frac{t}{2}, t+\frac{t}{2}\right)=\left(\frac{t}{2},\frac{3t}{2 }\right)$.
Where in the H did you get that. It is wrong.
$\left(t-\frac{t}{2}, t+\frac{t}{2}\right)=\{x: |x-t|<\frac{t}{2}\}\subset (-\infty,0)$.
You can also suppose that $L$ is a limit point of $[0,\infty)$ and show that $L\notin (-\infty,0)$

9. Re: Proving interval is closed in R Originally Posted by Plato Where in the H did you get that. It is wrong.
$\left(t-\frac{t}{2}, t+\frac{t}{2}\right)=\{x: |x-t|<\frac{t}{2}\}\subset (-\infty,0)$.
You can also suppose that $L$ is a limit point of $[0,\infty)$ and show that $L\notin (-\infty,0)$
I didn't know how to proceed. I just combined the like terms $\displaystyle t-\frac{t}{2}=\frac{t}{2}$ and $\displaystyle t+\frac{t}{2}=\frac{3t}{2}$. I still don't see how we prove $\displaystyle \left(t-\frac{t}{2}, t+\frac{t}{2}\right)\subset (-\infty,0)$