Show sup(A) is in F, where F is closed and open












0












$begingroup$


This question is worded very strangely and honestly I'm rather confused.



Suppose $varnothing$ $subsetneq$ F $subsetneq$ $mathbb{R}$ is closed. Then $exists$ x$_{0}$ $in$ F and y$_{0}$ $in$ F$^{c}$.



We will assume that x$_{0}$<y$_{0}$ without loss of generality.



The set A:={x $in$ F: x<y$_{0}$} is bounded above by y$_{0}$ and nonempty since x$_{0}$ $in$ A. As such, z:= sup(A) exists.



Show z $in$ F.



My professor mentioned that this had something to do with proving that $varnothing$ and $mathbb{R}$ are the only "clopen" subsets of $mathbb{R}$, but that's not actually mentioned anywhere in the question.






real-analysis metric-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your professor is wrong. A topological space has only two "clopen" is equivalent to say the topological space is connected (this is a tautology).
    $endgroup$
    – Will M.
    Nov 29 '18 at 2:14










  • $begingroup$
    @WillM. But $mathbb R$ is connected (under the Euclidean metric). And it's only clopen sets are $emptyset$ and $mathbb R$
    $endgroup$
    – fleablood
    Nov 29 '18 at 2:21










  • $begingroup$
    The result doesn't seem to be related right away with connectedness. I wouldn't know how to prove it is not related.
    $endgroup$
    – Will M.
    Nov 29 '18 at 2:55
















0












$begingroup$


This question is worded very strangely and honestly I'm rather confused.



Suppose $varnothing$ $subsetneq$ F $subsetneq$ $mathbb{R}$ is closed. Then $exists$ x$_{0}$ $in$ F and y$_{0}$ $in$ F$^{c}$.



We will assume that x$_{0}$<y$_{0}$ without loss of generality.



The set A:={x $in$ F: x<y$_{0}$} is bounded above by y$_{0}$ and nonempty since x$_{0}$ $in$ A. As such, z:= sup(A) exists.



Show z $in$ F.



My professor mentioned that this had something to do with proving that $varnothing$ and $mathbb{R}$ are the only "clopen" subsets of $mathbb{R}$, but that's not actually mentioned anywhere in the question.






real-analysis metric-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your professor is wrong. A topological space has only two "clopen" is equivalent to say the topological space is connected (this is a tautology).
    $endgroup$
    – Will M.
    Nov 29 '18 at 2:14










  • $begingroup$
    @WillM. But $mathbb R$ is connected (under the Euclidean metric). And it's only clopen sets are $emptyset$ and $mathbb R$
    $endgroup$
    – fleablood
    Nov 29 '18 at 2:21










  • $begingroup$
    The result doesn't seem to be related right away with connectedness. I wouldn't know how to prove it is not related.
    $endgroup$
    – Will M.
    Nov 29 '18 at 2:55














0












0








0





$begingroup$


This question is worded very strangely and honestly I'm rather confused.



Suppose $varnothing$ $subsetneq$ F $subsetneq$ $mathbb{R}$ is closed. Then $exists$ x$_{0}$ $in$ F and y$_{0}$ $in$ F$^{c}$.



We will assume that x$_{0}$<y$_{0}$ without loss of generality.



The set A:={x $in$ F: x<y$_{0}$} is bounded above by y$_{0}$ and nonempty since x$_{0}$ $in$ A. As such, z:= sup(A) exists.



Show z $in$ F.



My professor mentioned that this had something to do with proving that $varnothing$ and $mathbb{R}$ are the only "clopen" subsets of $mathbb{R}$, but that's not actually mentioned anywhere in the question.






real-analysis metric-spaces







share|cite|improve this question









$endgroup$




This question is worded very strangely and honestly I'm rather confused.



Suppose $varnothing$ $subsetneq$ F $subsetneq$ $mathbb{R}$ is closed. Then $exists$ x$_{0}$ $in$ F and y$_{0}$ $in$ F$^{c}$.



We will assume that x$_{0}$<y$_{0}$ without loss of generality.



The set A:={x $in$ F: x<y$_{0}$} is bounded above by y$_{0}$ and nonempty since x$_{0}$ $in$ A. As such, z:= sup(A) exists.



Show z $in$ F.



My professor mentioned that this had something to do with proving that $varnothing$ and $mathbb{R}$ are the only "clopen" subsets of $mathbb{R}$, but that's not actually mentioned anywhere in the question.






real-analysis metric-spaces




real-analysis metric-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '18 at 2:08









kendalkendal

337




337












  • $begingroup$
    Your professor is wrong. A topological space has only two "clopen" is equivalent to say the topological space is connected (this is a tautology).
    $endgroup$
    – Will M.
    Nov 29 '18 at 2:14










  • $begingroup$
    @WillM. But $mathbb R$ is connected (under the Euclidean metric). And it's only clopen sets are $emptyset$ and $mathbb R$
    $endgroup$
    – fleablood
    Nov 29 '18 at 2:21










  • $begingroup$
    The result doesn't seem to be related right away with connectedness. I wouldn't know how to prove it is not related.
    $endgroup$
    – Will M.
    Nov 29 '18 at 2:55


















  • $begingroup$
    Your professor is wrong. A topological space has only two "clopen" is equivalent to say the topological space is connected (this is a tautology).
    $endgroup$
    – Will M.
    Nov 29 '18 at 2:14










  • $begingroup$
    @WillM. But $mathbb R$ is connected (under the Euclidean metric). And it's only clopen sets are $emptyset$ and $mathbb R$
    $endgroup$
    – fleablood
    Nov 29 '18 at 2:21










  • $begingroup$
    The result doesn't seem to be related right away with connectedness. I wouldn't know how to prove it is not related.
    $endgroup$
    – Will M.
    Nov 29 '18 at 2:55
















$begingroup$
Your professor is wrong. A topological space has only two "clopen" is equivalent to say the topological space is connected (this is a tautology).
$endgroup$
– Will M.
Nov 29 '18 at 2:14




$begingroup$
Your professor is wrong. A topological space has only two "clopen" is equivalent to say the topological space is connected (this is a tautology).
$endgroup$
– Will M.
Nov 29 '18 at 2:14












$begingroup$
@WillM. But $mathbb R$ is connected (under the Euclidean metric). And it's only clopen sets are $emptyset$ and $mathbb R$
$endgroup$
– fleablood
Nov 29 '18 at 2:21




$begingroup$
@WillM. But $mathbb R$ is connected (under the Euclidean metric). And it's only clopen sets are $emptyset$ and $mathbb R$
$endgroup$
– fleablood
Nov 29 '18 at 2:21












$begingroup$
The result doesn't seem to be related right away with connectedness. I wouldn't know how to prove it is not related.
$endgroup$
– Will M.
Nov 29 '18 at 2:55




$begingroup$
The result doesn't seem to be related right away with connectedness. I wouldn't know how to prove it is not related.
$endgroup$
– Will M.
Nov 29 '18 at 2:55










1 Answer
1






active

oldest

votes


















2












$begingroup$

If $z in F^c$ then $znot in F$ and $z not in A$. But $z$ is $sup A$ so for every $epsilon$ there is a $x in A subset F$ so that $z-epsilon < x le z$ but as $x in A$, $x ne z$ so $z-epsilon < x < z$.



So $z$ is a limit point of $F$. But $F$ is closed. So $z in F$. A contradiction.



So $zin F$.



Not sure that this anything to do with clopen sets though. At least not immediately.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We don't assume $y_0$ is an upper bound of $F$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:32










  • $begingroup$
    Neither did I. $y_0$ is an upper bound of $A subset F$.
    $endgroup$
    – fleablood
    Nov 29 '18 at 2:33










  • $begingroup$
    Sure, but I don't understand how you claim that $zgeq y_0$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:36










  • $begingroup$
    Because if $z not in F$ then $z not in A$.
    $endgroup$
    – fleablood
    Nov 29 '18 at 2:37










  • $begingroup$
    Ok, but if $F=[0,1]$, $y_0=100$, then $A=F$ and $z=1<100=y_0$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:39













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

If $z in F^c$ then $znot in F$ and $z not in A$. But $z$ is $sup A$ so for every $epsilon$ there is a $x in A subset F$ so that $z-epsilon < x le z$ but as $x in A$, $x ne z$ so $z-epsilon < x < z$.



So $z$ is a limit point of $F$. But $F$ is closed. So $z in F$. A contradiction.



So $zin F$.



Not sure that this anything to do with clopen sets though. At least not immediately.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We don't assume $y_0$ is an upper bound of $F$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:32










  • $begingroup$
    Neither did I. $y_0$ is an upper bound of $A subset F$.
    $endgroup$
    – fleablood
    Nov 29 '18 at 2:33










  • $begingroup$
    Sure, but I don't understand how you claim that $zgeq y_0$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:36










  • $begingroup$
    Because if $z not in F$ then $z not in A$.
    $endgroup$
    – fleablood
    Nov 29 '18 at 2:37










  • $begingroup$
    Ok, but if $F=[0,1]$, $y_0=100$, then $A=F$ and $z=1<100=y_0$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:39


















2












$begingroup$

If $z in F^c$ then $znot in F$ and $z not in A$. But $z$ is $sup A$ so for every $epsilon$ there is a $x in A subset F$ so that $z-epsilon < x le z$ but as $x in A$, $x ne z$ so $z-epsilon < x < z$.



So $z$ is a limit point of $F$. But $F$ is closed. So $z in F$. A contradiction.



So $zin F$.



Not sure that this anything to do with clopen sets though. At least not immediately.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    We don't assume $y_0$ is an upper bound of $F$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:32










  • $begingroup$
    Neither did I. $y_0$ is an upper bound of $A subset F$.
    $endgroup$
    – fleablood
    Nov 29 '18 at 2:33










  • $begingroup$
    Sure, but I don't understand how you claim that $zgeq y_0$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:36










  • $begingroup$
    Because if $z not in F$ then $z not in A$.
    $endgroup$
    – fleablood
    Nov 29 '18 at 2:37










  • $begingroup$
    Ok, but if $F=[0,1]$, $y_0=100$, then $A=F$ and $z=1<100=y_0$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:39
















2












2








2





$begingroup$

If $z in F^c$ then $znot in F$ and $z not in A$. But $z$ is $sup A$ so for every $epsilon$ there is a $x in A subset F$ so that $z-epsilon < x le z$ but as $x in A$, $x ne z$ so $z-epsilon < x < z$.



So $z$ is a limit point of $F$. But $F$ is closed. So $z in F$. A contradiction.



So $zin F$.



Not sure that this anything to do with clopen sets though. At least not immediately.






share|cite|improve this answer











$endgroup$



If $z in F^c$ then $znot in F$ and $z not in A$. But $z$ is $sup A$ so for every $epsilon$ there is a $x in A subset F$ so that $z-epsilon < x le z$ but as $x in A$, $x ne z$ so $z-epsilon < x < z$.



So $z$ is a limit point of $F$. But $F$ is closed. So $z in F$. A contradiction.



So $zin F$.



Not sure that this anything to do with clopen sets though. At least not immediately.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '18 at 2:42

























answered Nov 29 '18 at 2:31









fleabloodfleablood

69.8k22685




69.8k22685












  • $begingroup$
    We don't assume $y_0$ is an upper bound of $F$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:32










  • $begingroup$
    Neither did I. $y_0$ is an upper bound of $A subset F$.
    $endgroup$
    – fleablood
    Nov 29 '18 at 2:33










  • $begingroup$
    Sure, but I don't understand how you claim that $zgeq y_0$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:36










  • $begingroup$
    Because if $z not in F$ then $z not in A$.
    $endgroup$
    – fleablood
    Nov 29 '18 at 2:37










  • $begingroup$
    Ok, but if $F=[0,1]$, $y_0=100$, then $A=F$ and $z=1<100=y_0$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:39




















  • $begingroup$
    We don't assume $y_0$ is an upper bound of $F$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:32










  • $begingroup$
    Neither did I. $y_0$ is an upper bound of $A subset F$.
    $endgroup$
    – fleablood
    Nov 29 '18 at 2:33










  • $begingroup$
    Sure, but I don't understand how you claim that $zgeq y_0$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:36










  • $begingroup$
    Because if $z not in F$ then $z not in A$.
    $endgroup$
    – fleablood
    Nov 29 '18 at 2:37










  • $begingroup$
    Ok, but if $F=[0,1]$, $y_0=100$, then $A=F$ and $z=1<100=y_0$.
    $endgroup$
    – Aweygan
    Nov 29 '18 at 2:39


















$begingroup$
We don't assume $y_0$ is an upper bound of $F$.
$endgroup$
– Aweygan
Nov 29 '18 at 2:32




$begingroup$
We don't assume $y_0$ is an upper bound of $F$.
$endgroup$
– Aweygan
Nov 29 '18 at 2:32












$begingroup$
Neither did I. $y_0$ is an upper bound of $A subset F$.
$endgroup$
– fleablood
Nov 29 '18 at 2:33




$begingroup$
Neither did I. $y_0$ is an upper bound of $A subset F$.
$endgroup$
– fleablood
Nov 29 '18 at 2:33












$begingroup$
Sure, but I don't understand how you claim that $zgeq y_0$.
$endgroup$
– Aweygan
Nov 29 '18 at 2:36




$begingroup$
Sure, but I don't understand how you claim that $zgeq y_0$.
$endgroup$
– Aweygan
Nov 29 '18 at 2:36












$begingroup$
Because if $z not in F$ then $z not in A$.
$endgroup$
– fleablood
Nov 29 '18 at 2:37




$begingroup$
Because if $z not in F$ then $z not in A$.
$endgroup$
– fleablood
Nov 29 '18 at 2:37












$begingroup$
Ok, but if $F=[0,1]$, $y_0=100$, then $A=F$ and $z=1<100=y_0$.
$endgroup$
– Aweygan
Nov 29 '18 at 2:39






$begingroup$
Ok, but if $F=[0,1]$, $y_0=100$, then $A=F$ and $z=1<100=y_0$.
$endgroup$
– Aweygan
Nov 29 '18 at 2:39




















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