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About Michael Spivak's proof of Heine - Borel Theorem for the line in his book “Calculus on Manifoldsâ€
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I cannot understand the following Michael Spivak's proof of Heine - Borel Theorem for the line.
I think that we must show $a < alpha$ first of all.
But he didn't prove that $a < alpha$.
Is his proof correct?
The closed interval $[a, b]$ is compact.
Proof.
If $mathcalO$ is an open cover of $[a, b]$,
let $A = x : a leq x leq b text and [a, x] text is covered by some finite number of open sets in mathcalO.$
Note that $a in A$ and that $A$ is clearly bounded above (by $b$).
We would like to show that $b in A$.
This is done by proving two things about $alpha =$ least upper bound of $A$ ; namely,
(1) $alpha in A$ and
(2) $b = alpha$.
Since $mathcalO$ is a cover, $alpha in U$ for some $U$ in $mathcalO$.
Then all points in some interval to the left of $alpha$ are also in $U$.
Since $alpha$ is the least upper bound of $A$, there is an $x$ in this interval such that $x in A$.
calculus general-topology compactness
asked yesterday
tchappy ha
1257
add a comment |Â
up vote
0
down vote
favorite
I cannot understand the following Michael Spivak's proof of Heine - Borel Theorem for the line.
I think that we must show $a < alpha$ first of all.
But he didn't prove that $a < alpha$.
Is his proof correct?
The closed interval $[a, b]$ is compact.
Proof.
If $mathcalO$ is an open cover of $[a, b]$,
let $A = x : a leq x leq b text and [a, x] text is covered by some finite number of open sets in mathcalO.$
Note that $a in A$ and that $A$ is clearly bounded above (by $b$).
We would like to show that $b in A$.
This is done by proving two things about $alpha =$ least upper bound of $A$ ; namely,
(1) $alpha in A$ and
(2) $b = alpha$.
Since $mathcalO$ is a cover, $alpha in U$ for some $U$ in $mathcalO$.
Then all points in some interval to the left of $alpha$ are also in $U$.
Since $alpha$ is the least upper bound of $A$, there is an $x$ in this interval such that $x in A$.
calculus general-topology compactness
asked yesterday
tchappy ha
1257
Michael Spivak says that there is an $x in A$. But why?
– tchappy ha
yesterday
1
Clearly $ain A$.
– quasi
yesterday
1
Also, the elements of the interval that cover $a$ are also in $A$, hence in fact, $alpha > a$.
– quasi
yesterday
Michael Spivak says that there is an $x in A$ such that $x < a$. But why?
– tchappy ha
yesterday
Sorry, I mean $x < alpha$
– tchappy ha
yesterday
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I cannot understand the following Michael Spivak's proof of Heine - Borel Theorem for the line.
I think that we must show $a < alpha$ first of all.
But he didn't prove that $a < alpha$.
Is his proof correct?
The closed interval $[a, b]$ is compact.
Proof.
If $mathcalO$ is an open cover of $[a, b]$,
let $A = x : a leq x leq b text and [a, x] text is covered by some finite number of open sets in mathcalO.$
Note that $a in A$ and that $A$ is clearly bounded above (by $b$).
We would like to show that $b in A$.
This is done by proving two things about $alpha =$ least upper bound of $A$ ; namely,
(1) $alpha in A$ and
(2) $b = alpha$.
Since $mathcalO$ is a cover, $alpha in U$ for some $U$ in $mathcalO$.
Then all points in some interval to the left of $alpha$ are also in $U$.
Since $alpha$ is the least upper bound of $A$, there is an $x$ in this interval such that $x in A$.
calculus general-topology compactness
asked yesterday
tchappy ha
1257
I cannot understand the following Michael Spivak's proof of Heine - Borel Theorem for the line.
I think that we must show $a < alpha$ first of all.
But he didn't prove that $a < alpha$.
Is his proof correct?
The closed interval $[a, b]$ is compact.
Proof.
If $mathcalO$ is an open cover of $[a, b]$,
let $A = x : a leq x leq b text and [a, x] text is covered by some finite number of open sets in mathcalO.$
Note that $a in A$ and that $A$ is clearly bounded above (by $b$).
We would like to show that $b in A$.
This is done by proving two things about $alpha =$ least upper bound of $A$ ; namely,
(1) $alpha in A$ and
(2) $b = alpha$.
Since $mathcalO$ is a cover, $alpha in U$ for some $U$ in $mathcalO$.
Then all points in some interval to the left of $alpha$ are also in $U$.
Since $alpha$ is the least upper bound of $A$, there is an $x$ in this interval such that $x in A$.
calculus general-topology compactness
asked yesterday
tchappy ha
1257
edited yesterday
asked yesterday
tchappy ha
1257
asked yesterday
tchappy ha
1257
asked yesterday
tchappy ha
1257
1257
Michael Spivak says that there is an $x in A$. But why?
– tchappy ha
yesterday
1
Clearly $ain A$.
– quasi
yesterday
1
Also, the elements of the interval that cover $a$ are also in $A$, hence in fact, $alpha > a$.
– quasi
yesterday
Michael Spivak says that there is an $x in A$ such that $x < a$. But why?
– tchappy ha
yesterday
Sorry, I mean $x < alpha$
– tchappy ha
yesterday
add a comment |Â
Michael Spivak says that there is an $x in A$. But why?
– tchappy ha
yesterday
1
Clearly $ain A$.
– quasi
yesterday
1
Also, the elements of the interval that cover $a$ are also in $A$, hence in fact, $alpha > a$.
– quasi
yesterday
Michael Spivak says that there is an $x in A$ such that $x < a$. But why?
– tchappy ha
yesterday
Sorry, I mean $x < alpha$
– tchappy ha
yesterday
Michael Spivak says that there is an $x in A$. But why?
– tchappy ha
yesterday
Michael Spivak says that there is an $x in A$. But why?
– tchappy ha
yesterday
1
1
Clearly $ain A$.
– quasi
yesterday
Clearly $ain A$.
– quasi
yesterday
1
1
Also, the elements of the interval that cover $a$ are also in $A$, hence in fact, $alpha > a$.
– quasi
yesterday
Also, the elements of the interval that cover $a$ are also in $A$, hence in fact, $alpha > a$.
– quasi
yesterday
Michael Spivak says that there is an $x in A$ such that $x < a$. But why?
– tchappy ha
yesterday
Michael Spivak says that there is an $x in A$ such that $x < a$. But why?
– tchappy ha
yesterday
Sorry, I mean $x < alpha$
– tchappy ha
yesterday
Sorry, I mean $x < alpha$
– tchappy ha
yesterday
add a comment |Â
2 Answers
2
active
oldest
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up vote
1
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accepted
Clearly $ain A$.
An element of one of the covering intervals is in $A$ if and only if all elements of that interval are in $A$.
In particular, the elements of an interval covering $a$ are in $A$.
Letting $alpha$ be the LUB of $A$, it follows that $alpha > a$.
Now consider an interval which covers $alpha$.
If $alphanotin A$, then all the elements of an interval covering $alpha$ which are slightly to the left of $alpha$ would not be in $A$, contrary the choice of $alpha$ as the LUB of $A$.
Hence $alphain A$.
But if $alpha < b$, then the elements of an interval covering $alpha$ which are slightly to the right of $alpha$ would be in $A$, again contrary the choice of $alpha$ as the LUB of $A$.
Hence $alpha = b$.
Thus, since $alphain A$, we get $bin A$, hence $A=[a,b]$.
answered yesterday
quasi
32.9k22258
If $U$ covers $[a, a]$, then $U$ covers all points in some interval to the right of $a$. So, $a < alpha$.
– tchappy ha
yesterday
1
Yes, I used that logic in my claim that $alpha > a$.
– quasi
yesterday
Thank you very much, quasi. I think I have understood the proof thanks to your answer and comment.
– tchappy ha
yesterday
add a comment |Â
up vote
0
down vote
My proof is the following:
The closed interval $[a, b]$ is compact.
Proof.
If $mathcalO$ is an open cover of $[a, b]$,
let $A = x : a leq x leq b text and [a, x] text is covered by some finite number of open sets in mathcalO.$
Note that $a in A$ and that $A$ is clearly bounded above (by $b$).
Since $mathcalO$ is a cover of $[a, b]$, $a in V$ for some $V$ in $mathcalO$.
There exists $delta > 0$ such that $a + delta leq b$ and $(a - delta, a + delta) in V$.
Let $a^'$ be a real number such that $a < a^' < a + delta leq b$.
Then $[a, a^'] subset (a - delta, a + delta) subset V$.
$therefore a^' in A$.
$therefore a < a^' leq alpha$.
$therefore a < alpha$.
Now we assume $alpha notin A$.
Since $mathcalO$ is a cover of $[a, b]$ and $alpha leq b$, $alpha in U$ for some $U$ in $mathcalO$.
Let $epsilon > 0$ be a real number such that $a < alpha - epsilon$ and $(alpha - epsilon, alpha + epsilon) in U$.
Then, there exists $beta$ such that $alpha - epsilon < beta leq alpha$ and $beta in A$ because $alpha$ is the least upper bound of $A$.
Since $alpha notin A$ by our assumption, $beta < alpha$.
$[a, beta]$ is covered by some finite number of open sets in $mathcalO$ and $[beta, alpha]$ is covered by $U$ in $mathcalO$.
So, $[a, alpha]$ is covered by some finite number of open sets in $mathcalO$.
$therefore alpha in A$.
But this is a contradiction to our assumption that $alpha notin A$.
$therefore alpha in A$.
Next, there exists $gamma$ such that $alpha < gamma$ and $gamma in (alpha - epsilon, alpha + epsilon) subset U$.
$[a, alpha]$ is covered by some finite number of open sets in $mathcalO$ and $[alpha, gamma]$ is covered by $U$ in $mathcalO$.
So, $[a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
If $gamma leq b$, then $gamma in A$.
This is a contradiction to the assumption that $alpha$ is an upper bound of $A$.
So $b < gamma$.
$[a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
So, of course $[a, b] subset [a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
answered 10 hours ago
tchappy ha
1257
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Clearly $ain A$.
An element of one of the covering intervals is in $A$ if and only if all elements of that interval are in $A$.
In particular, the elements of an interval covering $a$ are in $A$.
Letting $alpha$ be the LUB of $A$, it follows that $alpha > a$.
Now consider an interval which covers $alpha$.
If $alphanotin A$, then all the elements of an interval covering $alpha$ which are slightly to the left of $alpha$ would not be in $A$, contrary the choice of $alpha$ as the LUB of $A$.
Hence $alphain A$.
But if $alpha < b$, then the elements of an interval covering $alpha$ which are slightly to the right of $alpha$ would be in $A$, again contrary the choice of $alpha$ as the LUB of $A$.
Hence $alpha = b$.
Thus, since $alphain A$, we get $bin A$, hence $A=[a,b]$.
answered yesterday
quasi
32.9k22258
If $U$ covers $[a, a]$, then $U$ covers all points in some interval to the right of $a$. So, $a < alpha$.
– tchappy ha
yesterday
1
Yes, I used that logic in my claim that $alpha > a$.
– quasi
yesterday
Thank you very much, quasi. I think I have understood the proof thanks to your answer and comment.
– tchappy ha
yesterday
add a comment |Â
up vote
1
down vote
accepted
Clearly $ain A$.
An element of one of the covering intervals is in $A$ if and only if all elements of that interval are in $A$.
In particular, the elements of an interval covering $a$ are in $A$.
Letting $alpha$ be the LUB of $A$, it follows that $alpha > a$.
Now consider an interval which covers $alpha$.
If $alphanotin A$, then all the elements of an interval covering $alpha$ which are slightly to the left of $alpha$ would not be in $A$, contrary the choice of $alpha$ as the LUB of $A$.
Hence $alphain A$.
But if $alpha < b$, then the elements of an interval covering $alpha$ which are slightly to the right of $alpha$ would be in $A$, again contrary the choice of $alpha$ as the LUB of $A$.
Hence $alpha = b$.
Thus, since $alphain A$, we get $bin A$, hence $A=[a,b]$.
answered yesterday
quasi
32.9k22258
If $U$ covers $[a, a]$, then $U$ covers all points in some interval to the right of $a$. So, $a < alpha$.
– tchappy ha
yesterday
1
Yes, I used that logic in my claim that $alpha > a$.
– quasi
yesterday
Thank you very much, quasi. I think I have understood the proof thanks to your answer and comment.
– tchappy ha
yesterday
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Clearly $ain A$.
An element of one of the covering intervals is in $A$ if and only if all elements of that interval are in $A$.
In particular, the elements of an interval covering $a$ are in $A$.
Letting $alpha$ be the LUB of $A$, it follows that $alpha > a$.
Now consider an interval which covers $alpha$.
If $alphanotin A$, then all the elements of an interval covering $alpha$ which are slightly to the left of $alpha$ would not be in $A$, contrary the choice of $alpha$ as the LUB of $A$.
Hence $alphain A$.
But if $alpha < b$, then the elements of an interval covering $alpha$ which are slightly to the right of $alpha$ would be in $A$, again contrary the choice of $alpha$ as the LUB of $A$.
Hence $alpha = b$.
Thus, since $alphain A$, we get $bin A$, hence $A=[a,b]$.
answered yesterday
quasi
32.9k22258
Clearly $ain A$.
An element of one of the covering intervals is in $A$ if and only if all elements of that interval are in $A$.
In particular, the elements of an interval covering $a$ are in $A$.
Letting $alpha$ be the LUB of $A$, it follows that $alpha > a$.
Now consider an interval which covers $alpha$.
If $alphanotin A$, then all the elements of an interval covering $alpha$ which are slightly to the left of $alpha$ would not be in $A$, contrary the choice of $alpha$ as the LUB of $A$.
Hence $alphain A$.
But if $alpha < b$, then the elements of an interval covering $alpha$ which are slightly to the right of $alpha$ would be in $A$, again contrary the choice of $alpha$ as the LUB of $A$.
Hence $alpha = b$.
Thus, since $alphain A$, we get $bin A$, hence $A=[a,b]$.
answered yesterday
quasi
32.9k22258
edited yesterday
answered yesterday
quasi
32.9k22258
answered yesterday
quasi
32.9k22258
answered yesterday
quasi
32.9k22258
32.9k22258
If $U$ covers $[a, a]$, then $U$ covers all points in some interval to the right of $a$. So, $a < alpha$.
– tchappy ha
yesterday
1
Yes, I used that logic in my claim that $alpha > a$.
– quasi
yesterday
Thank you very much, quasi. I think I have understood the proof thanks to your answer and comment.
– tchappy ha
yesterday
add a comment |Â
If $U$ covers $[a, a]$, then $U$ covers all points in some interval to the right of $a$. So, $a < alpha$.
– tchappy ha
yesterday
1
Yes, I used that logic in my claim that $alpha > a$.
– quasi
yesterday
Thank you very much, quasi. I think I have understood the proof thanks to your answer and comment.
– tchappy ha
yesterday
If $U$ covers $[a, a]$, then $U$ covers all points in some interval to the right of $a$. So, $a < alpha$.
– tchappy ha
yesterday
If $U$ covers $[a, a]$, then $U$ covers all points in some interval to the right of $a$. So, $a < alpha$.
– tchappy ha
yesterday
1
1
Yes, I used that logic in my claim that $alpha > a$.
– quasi
yesterday
Yes, I used that logic in my claim that $alpha > a$.
– quasi
yesterday
Thank you very much, quasi. I think I have understood the proof thanks to your answer and comment.
– tchappy ha
yesterday
Thank you very much, quasi. I think I have understood the proof thanks to your answer and comment.
– tchappy ha
yesterday
add a comment |Â
up vote
0
down vote
My proof is the following:
The closed interval $[a, b]$ is compact.
Proof.
If $mathcalO$ is an open cover of $[a, b]$,
let $A = x : a leq x leq b text and [a, x] text is covered by some finite number of open sets in mathcalO.$
Note that $a in A$ and that $A$ is clearly bounded above (by $b$).
Since $mathcalO$ is a cover of $[a, b]$, $a in V$ for some $V$ in $mathcalO$.
There exists $delta > 0$ such that $a + delta leq b$ and $(a - delta, a + delta) in V$.
Let $a^'$ be a real number such that $a < a^' < a + delta leq b$.
Then $[a, a^'] subset (a - delta, a + delta) subset V$.
$therefore a^' in A$.
$therefore a < a^' leq alpha$.
$therefore a < alpha$.
Now we assume $alpha notin A$.
Since $mathcalO$ is a cover of $[a, b]$ and $alpha leq b$, $alpha in U$ for some $U$ in $mathcalO$.
Let $epsilon > 0$ be a real number such that $a < alpha - epsilon$ and $(alpha - epsilon, alpha + epsilon) in U$.
Then, there exists $beta$ such that $alpha - epsilon < beta leq alpha$ and $beta in A$ because $alpha$ is the least upper bound of $A$.
Since $alpha notin A$ by our assumption, $beta < alpha$.
$[a, beta]$ is covered by some finite number of open sets in $mathcalO$ and $[beta, alpha]$ is covered by $U$ in $mathcalO$.
So, $[a, alpha]$ is covered by some finite number of open sets in $mathcalO$.
$therefore alpha in A$.
But this is a contradiction to our assumption that $alpha notin A$.
$therefore alpha in A$.
Next, there exists $gamma$ such that $alpha < gamma$ and $gamma in (alpha - epsilon, alpha + epsilon) subset U$.
$[a, alpha]$ is covered by some finite number of open sets in $mathcalO$ and $[alpha, gamma]$ is covered by $U$ in $mathcalO$.
So, $[a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
If $gamma leq b$, then $gamma in A$.
This is a contradiction to the assumption that $alpha$ is an upper bound of $A$.
So $b < gamma$.
$[a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
So, of course $[a, b] subset [a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
answered 10 hours ago
tchappy ha
1257
add a comment |Â
up vote
0
down vote
My proof is the following:
The closed interval $[a, b]$ is compact.
Proof.
If $mathcalO$ is an open cover of $[a, b]$,
let $A = x : a leq x leq b text and [a, x] text is covered by some finite number of open sets in mathcalO.$
Note that $a in A$ and that $A$ is clearly bounded above (by $b$).
Since $mathcalO$ is a cover of $[a, b]$, $a in V$ for some $V$ in $mathcalO$.
There exists $delta > 0$ such that $a + delta leq b$ and $(a - delta, a + delta) in V$.
Let $a^'$ be a real number such that $a < a^' < a + delta leq b$.
Then $[a, a^'] subset (a - delta, a + delta) subset V$.
$therefore a^' in A$.
$therefore a < a^' leq alpha$.
$therefore a < alpha$.
Now we assume $alpha notin A$.
Since $mathcalO$ is a cover of $[a, b]$ and $alpha leq b$, $alpha in U$ for some $U$ in $mathcalO$.
Let $epsilon > 0$ be a real number such that $a < alpha - epsilon$ and $(alpha - epsilon, alpha + epsilon) in U$.
Then, there exists $beta$ such that $alpha - epsilon < beta leq alpha$ and $beta in A$ because $alpha$ is the least upper bound of $A$.
Since $alpha notin A$ by our assumption, $beta < alpha$.
$[a, beta]$ is covered by some finite number of open sets in $mathcalO$ and $[beta, alpha]$ is covered by $U$ in $mathcalO$.
So, $[a, alpha]$ is covered by some finite number of open sets in $mathcalO$.
$therefore alpha in A$.
But this is a contradiction to our assumption that $alpha notin A$.
$therefore alpha in A$.
Next, there exists $gamma$ such that $alpha < gamma$ and $gamma in (alpha - epsilon, alpha + epsilon) subset U$.
$[a, alpha]$ is covered by some finite number of open sets in $mathcalO$ and $[alpha, gamma]$ is covered by $U$ in $mathcalO$.
So, $[a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
If $gamma leq b$, then $gamma in A$.
This is a contradiction to the assumption that $alpha$ is an upper bound of $A$.
So $b < gamma$.
$[a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
So, of course $[a, b] subset [a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
answered 10 hours ago
tchappy ha
1257
add a comment |Â
up vote
0
down vote
up vote
0
down vote
My proof is the following:
The closed interval $[a, b]$ is compact.
Proof.
If $mathcalO$ is an open cover of $[a, b]$,
let $A = x : a leq x leq b text and [a, x] text is covered by some finite number of open sets in mathcalO.$
Note that $a in A$ and that $A$ is clearly bounded above (by $b$).
Since $mathcalO$ is a cover of $[a, b]$, $a in V$ for some $V$ in $mathcalO$.
There exists $delta > 0$ such that $a + delta leq b$ and $(a - delta, a + delta) in V$.
Let $a^'$ be a real number such that $a < a^' < a + delta leq b$.
Then $[a, a^'] subset (a - delta, a + delta) subset V$.
$therefore a^' in A$.
$therefore a < a^' leq alpha$.
$therefore a < alpha$.
Now we assume $alpha notin A$.
Since $mathcalO$ is a cover of $[a, b]$ and $alpha leq b$, $alpha in U$ for some $U$ in $mathcalO$.
Let $epsilon > 0$ be a real number such that $a < alpha - epsilon$ and $(alpha - epsilon, alpha + epsilon) in U$.
Then, there exists $beta$ such that $alpha - epsilon < beta leq alpha$ and $beta in A$ because $alpha$ is the least upper bound of $A$.
Since $alpha notin A$ by our assumption, $beta < alpha$.
$[a, beta]$ is covered by some finite number of open sets in $mathcalO$ and $[beta, alpha]$ is covered by $U$ in $mathcalO$.
So, $[a, alpha]$ is covered by some finite number of open sets in $mathcalO$.
$therefore alpha in A$.
But this is a contradiction to our assumption that $alpha notin A$.
$therefore alpha in A$.
Next, there exists $gamma$ such that $alpha < gamma$ and $gamma in (alpha - epsilon, alpha + epsilon) subset U$.
$[a, alpha]$ is covered by some finite number of open sets in $mathcalO$ and $[alpha, gamma]$ is covered by $U$ in $mathcalO$.
So, $[a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
If $gamma leq b$, then $gamma in A$.
This is a contradiction to the assumption that $alpha$ is an upper bound of $A$.
So $b < gamma$.
$[a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
So, of course $[a, b] subset [a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
answered 10 hours ago
tchappy ha
1257
My proof is the following:
The closed interval $[a, b]$ is compact.
Proof.
If $mathcalO$ is an open cover of $[a, b]$,
let $A = x : a leq x leq b text and [a, x] text is covered by some finite number of open sets in mathcalO.$
Note that $a in A$ and that $A$ is clearly bounded above (by $b$).
Since $mathcalO$ is a cover of $[a, b]$, $a in V$ for some $V$ in $mathcalO$.
There exists $delta > 0$ such that $a + delta leq b$ and $(a - delta, a + delta) in V$.
Let $a^'$ be a real number such that $a < a^' < a + delta leq b$.
Then $[a, a^'] subset (a - delta, a + delta) subset V$.
$therefore a^' in A$.
$therefore a < a^' leq alpha$.
$therefore a < alpha$.
Now we assume $alpha notin A$.
Since $mathcalO$ is a cover of $[a, b]$ and $alpha leq b$, $alpha in U$ for some $U$ in $mathcalO$.
Let $epsilon > 0$ be a real number such that $a < alpha - epsilon$ and $(alpha - epsilon, alpha + epsilon) in U$.
Then, there exists $beta$ such that $alpha - epsilon < beta leq alpha$ and $beta in A$ because $alpha$ is the least upper bound of $A$.
Since $alpha notin A$ by our assumption, $beta < alpha$.
$[a, beta]$ is covered by some finite number of open sets in $mathcalO$ and $[beta, alpha]$ is covered by $U$ in $mathcalO$.
So, $[a, alpha]$ is covered by some finite number of open sets in $mathcalO$.
$therefore alpha in A$.
But this is a contradiction to our assumption that $alpha notin A$.
$therefore alpha in A$.
Next, there exists $gamma$ such that $alpha < gamma$ and $gamma in (alpha - epsilon, alpha + epsilon) subset U$.
$[a, alpha]$ is covered by some finite number of open sets in $mathcalO$ and $[alpha, gamma]$ is covered by $U$ in $mathcalO$.
So, $[a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
If $gamma leq b$, then $gamma in A$.
This is a contradiction to the assumption that $alpha$ is an upper bound of $A$.
So $b < gamma$.
$[a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
So, of course $[a, b] subset [a, gamma]$ is covered by some finite number of open sets in $mathcalO$.
answered 10 hours ago
tchappy ha
1257
edited 9 hours ago
answered 10 hours ago
tchappy ha
1257
answered 10 hours ago
tchappy ha
1257
answered 10 hours ago
tchappy ha
1257
1257
add a comment |Â
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Michael Spivak says that there is an $x in A$. But why?
– tchappy ha
yesterday
1
Clearly $ain A$.
– quasi
yesterday
1
Also, the elements of the interval that cover $a$ are also in $A$, hence in fact, $alpha > a$.
– quasi
yesterday
Michael Spivak says that there is an $x in A$ such that $x < a$. But why?
– tchappy ha
yesterday
Sorry, I mean $x < alpha$
– tchappy ha
yesterday