Show sup(A) is in F, where F is closed and open
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
asked Nov 29 '18 at 2:08
kendalkendal
337
add a comment |
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
asked Nov 29 '18 at 2:08
kendalkendal
337
Your professor is wrong. A topological space has only two "clopen" is equivalent to say the topological space is connected (this is a tautology).
– Will M.
Nov 29 '18 at 2:14
@WillM. But $mathbb R$ is connected (under the Euclidean metric). And it's only clopen sets are $emptyset$ and $mathbb R$
– fleablood
Nov 29 '18 at 2:21
The result doesn't seem to be related right away with connectedness. I wouldn't know how to prove it is not related.
– Will M.
Nov 29 '18 at 2:55
add a comment |
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
asked Nov 29 '18 at 2:08
kendalkendal
337
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
asked Nov 29 '18 at 2:08
kendalkendal
337
asked Nov 29 '18 at 2:08
kendalkendal
337
asked Nov 29 '18 at 2:08
kendalkendal
337
asked Nov 29 '18 at 2:08
kendalkendal
337
asked Nov 29 '18 at 2:08
kendalkendal
337
337
Your professor is wrong. A topological space has only two "clopen" is equivalent to say the topological space is connected (this is a tautology).
– Will M.
Nov 29 '18 at 2:14
@WillM. But $mathbb R$ is connected (under the Euclidean metric). And it's only clopen sets are $emptyset$ and $mathbb R$
– fleablood
Nov 29 '18 at 2:21
The result doesn't seem to be related right away with connectedness. I wouldn't know how to prove it is not related.
– Will M.
Nov 29 '18 at 2:55
add a comment |
Your professor is wrong. A topological space has only two "clopen" is equivalent to say the topological space is connected (this is a tautology).
– Will M.
Nov 29 '18 at 2:14
@WillM. But $mathbb R$ is connected (under the Euclidean metric). And it's only clopen sets are $emptyset$ and $mathbb R$
– fleablood
Nov 29 '18 at 2:21
The result doesn't seem to be related right away with connectedness. I wouldn't know how to prove it is not related.
– Will M.
Nov 29 '18 at 2:55
Your professor is wrong. A topological space has only two "clopen" is equivalent to say the topological space is connected (this is a tautology).
– Will M.
Nov 29 '18 at 2:14
Your professor is wrong. A topological space has only two "clopen" is equivalent to say the topological space is connected (this is a tautology).
– Will M.
Nov 29 '18 at 2:14
@WillM. But $mathbb R$ is connected (under the Euclidean metric). And it's only clopen sets are $emptyset$ and $mathbb R$
– fleablood
Nov 29 '18 at 2:21
@WillM. But $mathbb R$ is connected (under the Euclidean metric). And it's only clopen sets are $emptyset$ and $mathbb R$
– fleablood
Nov 29 '18 at 2:21
The result doesn't seem to be related right away with connectedness. I wouldn't know how to prove it is not related.
– Will M.
Nov 29 '18 at 2:55
The result doesn't seem to be related right away with connectedness. I wouldn't know how to prove it is not related.
– Will M.
Nov 29 '18 at 2:55
add a comment |
1 Answer
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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.
answered Nov 29 '18 at 2:31
fleabloodfleablood
68.6k22685
We don't assume $y_0$ is an upper bound of $F$.
– Aweygan
Nov 29 '18 at 2:32
Neither did I. $y_0$ is an upper bound of $A subset F$.
– fleablood
Nov 29 '18 at 2:33
Sure, but I don't understand how you claim that $zgeq y_0$.
– Aweygan
Nov 29 '18 at 2:36
Because if $z not in F$ then $z not in A$.
– fleablood
Nov 29 '18 at 2:37
Ok, but if $F=[0,1]$, $y_0=100$, then $A=F$ and $z=1<100=y_0$.
– Aweygan
Nov 29 '18 at 2:39
|
show 3 more comments
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1 Answer
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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.
answered Nov 29 '18 at 2:31
fleabloodfleablood
68.6k22685
We don't assume $y_0$ is an upper bound of $F$.
– Aweygan
Nov 29 '18 at 2:32
Neither did I. $y_0$ is an upper bound of $A subset F$.
– fleablood
Nov 29 '18 at 2:33
Sure, but I don't understand how you claim that $zgeq y_0$.
– Aweygan
Nov 29 '18 at 2:36
Because if $z not in F$ then $z not in A$.
– fleablood
Nov 29 '18 at 2:37
Ok, but if $F=[0,1]$, $y_0=100$, then $A=F$ and $z=1<100=y_0$.
– Aweygan
Nov 29 '18 at 2:39
|
show 3 more comments
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.
answered Nov 29 '18 at 2:31
fleabloodfleablood
68.6k22685
We don't assume $y_0$ is an upper bound of $F$.
– Aweygan
Nov 29 '18 at 2:32
Neither did I. $y_0$ is an upper bound of $A subset F$.
– fleablood
Nov 29 '18 at 2:33
Sure, but I don't understand how you claim that $zgeq y_0$.
– Aweygan
Nov 29 '18 at 2:36
Because if $z not in F$ then $z not in A$.
– fleablood
Nov 29 '18 at 2:37
Ok, but if $F=[0,1]$, $y_0=100$, then $A=F$ and $z=1<100=y_0$.
– Aweygan
Nov 29 '18 at 2:39
|
show 3 more comments
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.
answered Nov 29 '18 at 2:31
fleabloodfleablood
68.6k22685
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.
answered Nov 29 '18 at 2:31
fleabloodfleablood
68.6k22685
edited Nov 29 '18 at 2:42
answered Nov 29 '18 at 2:31
fleabloodfleablood
68.6k22685
answered Nov 29 '18 at 2:31
fleabloodfleablood
68.6k22685
answered Nov 29 '18 at 2:31
fleabloodfleablood
68.6k22685
68.6k22685
We don't assume $y_0$ is an upper bound of $F$.
– Aweygan
Nov 29 '18 at 2:32
Neither did I. $y_0$ is an upper bound of $A subset F$.
– fleablood
Nov 29 '18 at 2:33
Sure, but I don't understand how you claim that $zgeq y_0$.
– Aweygan
Nov 29 '18 at 2:36
Because if $z not in F$ then $z not in A$.
– fleablood
Nov 29 '18 at 2:37
Ok, but if $F=[0,1]$, $y_0=100$, then $A=F$ and $z=1<100=y_0$.
– Aweygan
Nov 29 '18 at 2:39
|
show 3 more comments
We don't assume $y_0$ is an upper bound of $F$.
– Aweygan
Nov 29 '18 at 2:32
Neither did I. $y_0$ is an upper bound of $A subset F$.
– fleablood
Nov 29 '18 at 2:33
Sure, but I don't understand how you claim that $zgeq y_0$.
– Aweygan
Nov 29 '18 at 2:36
Because if $z not in F$ then $z not in A$.
– fleablood
Nov 29 '18 at 2:37
Ok, but if $F=[0,1]$, $y_0=100$, then $A=F$ and $z=1<100=y_0$.
– Aweygan
Nov 29 '18 at 2:39
We don't assume $y_0$ is an upper bound of $F$.
– Aweygan
Nov 29 '18 at 2:32
We don't assume $y_0$ is an upper bound of $F$.
– Aweygan
Nov 29 '18 at 2:32
Neither did I. $y_0$ is an upper bound of $A subset F$.
– fleablood
Nov 29 '18 at 2:33
Neither did I. $y_0$ is an upper bound of $A subset F$.
– fleablood
Nov 29 '18 at 2:33
Sure, but I don't understand how you claim that $zgeq y_0$.
– Aweygan
Nov 29 '18 at 2:36
Sure, but I don't understand how you claim that $zgeq y_0$.
– Aweygan
Nov 29 '18 at 2:36
Because if $z not in F$ then $z not in A$.
– fleablood
Nov 29 '18 at 2:37
Because if $z not in F$ then $z not in A$.
– fleablood
Nov 29 '18 at 2:37
Ok, but if $F=[0,1]$, $y_0=100$, then $A=F$ and $z=1<100=y_0$.
– Aweygan
Nov 29 '18 at 2:39
Ok, but if $F=[0,1]$, $y_0=100$, then $A=F$ and $z=1<100=y_0$.
– Aweygan
Nov 29 '18 at 2:39
|
show 3 more comments
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Your professor is wrong. A topological space has only two "clopen" is equivalent to say the topological space is connected (this is a tautology).
– Will M.
Nov 29 '18 at 2:14
@WillM. But $mathbb R$ is connected (under the Euclidean metric). And it's only clopen sets are $emptyset$ and $mathbb R$
– fleablood
Nov 29 '18 at 2:21
The result doesn't seem to be related right away with connectedness. I wouldn't know how to prove it is not related.
– Will M.
Nov 29 '18 at 2:55