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Thread: Proving interval is closed in R

  1. #1
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    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?
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    Re: Proving interval is closed in R

    Quote Originally Posted by adkinsjr View Post
    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?
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    Re: Proving interval is closed in R

    Yes, to be clear the goal is to prove $\displaystyle [0,\infty)$ closed.
    Last edited by adkinsjr; Sep 13th 2019 at 12:57 PM.
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    Re: Proving interval is closed in R

    Quote Originally Posted by adkinsjr View Post
    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?
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    Re: Proving interval is closed in R

    Quote Originally Posted by Plato View Post
    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.
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    Re: Proving interval is closed in R

    Quote Originally Posted by adkinsjr View Post
    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 $[0,\infty)~?$
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    Re: Proving interval is closed in R

    Quote Originally Posted by Plato View Post
    O.K. Suppose that $t\in(-\infty,0)$ then $|t|>0$ Why is that true?
    Absolute values are always greater than 0.

    Quote Originally Posted by Plato View Post
    Define $\delta=\frac{|t|}{2}$ Why is $\delta>0$?
    Because $\displaystyle |t|>0$

    Quote Originally Posted by Plato View Post
    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.
    Last edited by adkinsjr; Sep 14th 2019 at 12:29 PM.
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    Re: Proving interval is closed in R

    Quote Originally Posted by adkinsjr View Post
    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 $L\notin (-\infty,0)$
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    Re: Proving interval is closed in R

    Quote Originally Posted by Plato View Post
    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 $\displaystyle \left(t-\frac{t}{2}, t+\frac{t}{2}\right)\subset (-\infty,0)$
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