let $xinemptysetsubseteqmathbb{R}$ why is $xleq r,forall r inmathbb{R}$ true
$begingroup$
The emtyset has no element one can compare it with so why the statement is true can somebody explain the logic of this. I ask because I have seen a post where one claims that $sup(emptyset)=-infty$, the argument was that the above statement is always true and if there would exist a $rinmathbb{R}$ then one can always find a smaller upperbound. Id on't see why this argument is always true and hope somebody can explain it to me.
For reference
Why is the supremum of the empty set $-infty$ and the infimum $infty$?
In particular the answer of Clive Newstead
real-analysis supremum-and-infimum
asked Jan 24 at 16:27
New2MathNew2Math
8912
$endgroup$
add a comment |
$begingroup$
The emtyset has no element one can compare it with so why the statement is true can somebody explain the logic of this. I ask because I have seen a post where one claims that $sup(emptyset)=-infty$, the argument was that the above statement is always true and if there would exist a $rinmathbb{R}$ then one can always find a smaller upperbound. Id on't see why this argument is always true and hope somebody can explain it to me.
For reference
Why is the supremum of the empty set $-infty$ and the infimum $infty$?
In particular the answer of Clive Newstead
real-analysis supremum-and-infimum
asked Jan 24 at 16:27
New2MathNew2Math
8912
$endgroup$
1
$begingroup$
You have to consider the condition : "if $x in emptyset$, then $x le r$", for an $r in mathbb R$ whatever.
$endgroup$
– Mauro ALLEGRANZA
Jan 24 at 16:33
1
$begingroup$
Due to the fact that $x in emptyset$ is always false the above condition is always true, for every $r in mathbb R$, whatever small $r$ is.
$endgroup$
– Mauro ALLEGRANZA
Jan 24 at 16:34
add a comment |
$begingroup$
The emtyset has no element one can compare it with so why the statement is true can somebody explain the logic of this. I ask because I have seen a post where one claims that $sup(emptyset)=-infty$, the argument was that the above statement is always true and if there would exist a $rinmathbb{R}$ then one can always find a smaller upperbound. Id on't see why this argument is always true and hope somebody can explain it to me.
For reference
Why is the supremum of the empty set $-infty$ and the infimum $infty$?
In particular the answer of Clive Newstead
real-analysis supremum-and-infimum
asked Jan 24 at 16:27
New2MathNew2Math
8912
$endgroup$
The emtyset has no element one can compare it with so why the statement is true can somebody explain the logic of this. I ask because I have seen a post where one claims that $sup(emptyset)=-infty$, the argument was that the above statement is always true and if there would exist a $rinmathbb{R}$ then one can always find a smaller upperbound. Id on't see why this argument is always true and hope somebody can explain it to me.
For reference
Why is the supremum of the empty set $-infty$ and the infimum $infty$?
In particular the answer of Clive Newstead
real-analysis supremum-and-infimum
real-analysis supremum-and-infimum
asked Jan 24 at 16:27
New2MathNew2Math
8912
asked Jan 24 at 16:27
New2MathNew2Math
8912
edited Jan 24 at 16:43
New2Math
asked Jan 24 at 16:27
New2MathNew2Math
8912
asked Jan 24 at 16:27
New2MathNew2Math
8912
asked Jan 24 at 16:27
New2MathNew2Math
8912
8912
1
$begingroup$
You have to consider the condition : "if $x in emptyset$, then $x le r$", for an $r in mathbb R$ whatever.
$endgroup$
– Mauro ALLEGRANZA
Jan 24 at 16:33
1
$begingroup$
Due to the fact that $x in emptyset$ is always false the above condition is always true, for every $r in mathbb R$, whatever small $r$ is.
$endgroup$
– Mauro ALLEGRANZA
Jan 24 at 16:34
add a comment |
1
$begingroup$
You have to consider the condition : "if $x in emptyset$, then $x le r$", for an $r in mathbb R$ whatever.
$endgroup$
– Mauro ALLEGRANZA
Jan 24 at 16:33
1
$begingroup$
Due to the fact that $x in emptyset$ is always false the above condition is always true, for every $r in mathbb R$, whatever small $r$ is.
$endgroup$
– Mauro ALLEGRANZA
Jan 24 at 16:34
1
1
$begingroup$
You have to consider the condition : "if $x in emptyset$, then $x le r$", for an $r in mathbb R$ whatever.
$endgroup$
– Mauro ALLEGRANZA
Jan 24 at 16:33
$begingroup$
You have to consider the condition : "if $x in emptyset$, then $x le r$", for an $r in mathbb R$ whatever.
$endgroup$
– Mauro ALLEGRANZA
Jan 24 at 16:33
1
1
$begingroup$
Due to the fact that $x in emptyset$ is always false the above condition is always true, for every $r in mathbb R$, whatever small $r$ is.
$endgroup$
– Mauro ALLEGRANZA
Jan 24 at 16:34
$begingroup$
Due to the fact that $x in emptyset$ is always false the above condition is always true, for every $r in mathbb R$, whatever small $r$ is.
$endgroup$
– Mauro ALLEGRANZA
Jan 24 at 16:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Three explanations: (But they all involve that as there are no $x in emptyset$ we can not negate $x le r$ for any $r in mathbb R$)
....
Either $xin emptyset implies x le r :forall r in mathbb R$ is true or it is false.
If it is false then that implies that there is an $x in emptyset$ and an $r in mathbb R$ so that $x > r$. That is impossible as there are no $x in emptyset$.
If it is true then for every $x in emptyset$ then ... something. We can never test this because we can never find an $x in emptyset$.
However we can do logic the statement $P implies Q$ will be true if $P$ and $Q$ are both true, or if $P$ is false. It will only be false if $P$ is true and $Q$ is false. If $P= : x in emptyset$ and $Q = : x le r :forall r in mathbb R$ then $P$ is always false. And $Q$ can never be true for any actuall $x in mathbb R$. So $P implies Q$ is true.
Also a statement is equivalent to the contrapostive. The contrapositive of $xin emptyset implies x le r:forall r in mathbb R$ is:
$exists rin mathbb R: r < x implies xnot in emptyset$. Well that is certainly true! If $r < x$ then $x$ must exist. And if $x$ exists then $x not in emptyset$!.
......
If $xin emptyset$ then $x$ is a green cheese eating alien who is the reincarnation of Elvis Presley is true because logically a false premise implies all conclusions.
So if $x in emptyset$ then $x le r$ for all $r in mathbb R$. (it's also true that $x >r$ fora all $r in mathbb R$ and that $x = r$ for all $r in mathbb R$ and so on.... as $x$ does not exist we can say anything we want to be true about it.)
....
Or looking at it another way: If $A_r = (-infty, r)$ then $emptyset subset A_r$ because the empty set is a subset of all sets. (The emptyset has no elements, so no elements aren't in any set.) So if $x in emptyset implies x in A_r$.
That is true for all $A_r: rin mathbb R$. So $x in (-infty, r)$ for all $rin mathbb R$ so $x le r$ for all $r in mathbb R$.
You might say "there's got to be a trick in there somewhere; nothing can be in all of those intervals" and you'd be right. The trick is that no such $xin emptyset$ does exist. But if such an $x$ DID exist, it would have to be in every set including every interval including those intervals.
answered Jan 24 at 17:07
fleabloodfleablood
72k22687
$endgroup$
$begingroup$
Why is it logical to say : a false premise implies all conclusions?
$endgroup$
– New2Math
Jan 24 at 19:59
$begingroup$
because it is..... $P$ implies $Q$ can be proven two ways, i) by checking that every time $P$ is true so is $Q$. We can't perform that test because $P$ is never true. or ii) by checking there's never a time when $P$ is true but $Q$ is false. And it's obvious there is never a time when $P$ is true but $Q$ is false because there is never a time that $P$ is true.
$endgroup$
– fleablood
Jan 24 at 20:08
add a comment |
$begingroup$
Given $r$, every element of the empty set is at most $r$. This statement is true because its negation is false. Namely, its negation is that there exists $r$ and an element of the empty set that is at greater than $r$. But the empty set has no elements.
answered Jan 24 at 16:39
Foobaz JohnFoobaz John
22.5k41452
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
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$begingroup$
Three explanations: (But they all involve that as there are no $x in emptyset$ we can not negate $x le r$ for any $r in mathbb R$)
....
Either $xin emptyset implies x le r :forall r in mathbb R$ is true or it is false.
If it is false then that implies that there is an $x in emptyset$ and an $r in mathbb R$ so that $x > r$. That is impossible as there are no $x in emptyset$.
If it is true then for every $x in emptyset$ then ... something. We can never test this because we can never find an $x in emptyset$.
However we can do logic the statement $P implies Q$ will be true if $P$ and $Q$ are both true, or if $P$ is false. It will only be false if $P$ is true and $Q$ is false. If $P= : x in emptyset$ and $Q = : x le r :forall r in mathbb R$ then $P$ is always false. And $Q$ can never be true for any actuall $x in mathbb R$. So $P implies Q$ is true.
Also a statement is equivalent to the contrapostive. The contrapositive of $xin emptyset implies x le r:forall r in mathbb R$ is:
$exists rin mathbb R: r < x implies xnot in emptyset$. Well that is certainly true! If $r < x$ then $x$ must exist. And if $x$ exists then $x not in emptyset$!.
......
If $xin emptyset$ then $x$ is a green cheese eating alien who is the reincarnation of Elvis Presley is true because logically a false premise implies all conclusions.
So if $x in emptyset$ then $x le r$ for all $r in mathbb R$. (it's also true that $x >r$ fora all $r in mathbb R$ and that $x = r$ for all $r in mathbb R$ and so on.... as $x$ does not exist we can say anything we want to be true about it.)
....
Or looking at it another way: If $A_r = (-infty, r)$ then $emptyset subset A_r$ because the empty set is a subset of all sets. (The emptyset has no elements, so no elements aren't in any set.) So if $x in emptyset implies x in A_r$.
That is true for all $A_r: rin mathbb R$. So $x in (-infty, r)$ for all $rin mathbb R$ so $x le r$ for all $r in mathbb R$.
You might say "there's got to be a trick in there somewhere; nothing can be in all of those intervals" and you'd be right. The trick is that no such $xin emptyset$ does exist. But if such an $x$ DID exist, it would have to be in every set including every interval including those intervals.
answered Jan 24 at 17:07
fleabloodfleablood
72k22687
$endgroup$
$begingroup$
Why is it logical to say : a false premise implies all conclusions?
$endgroup$
– New2Math
Jan 24 at 19:59
$begingroup$
because it is..... $P$ implies $Q$ can be proven two ways, i) by checking that every time $P$ is true so is $Q$. We can't perform that test because $P$ is never true. or ii) by checking there's never a time when $P$ is true but $Q$ is false. And it's obvious there is never a time when $P$ is true but $Q$ is false because there is never a time that $P$ is true.
$endgroup$
– fleablood
Jan 24 at 20:08
add a comment |
$begingroup$
Three explanations: (But they all involve that as there are no $x in emptyset$ we can not negate $x le r$ for any $r in mathbb R$)
....
Either $xin emptyset implies x le r :forall r in mathbb R$ is true or it is false.
If it is false then that implies that there is an $x in emptyset$ and an $r in mathbb R$ so that $x > r$. That is impossible as there are no $x in emptyset$.
If it is true then for every $x in emptyset$ then ... something. We can never test this because we can never find an $x in emptyset$.
However we can do logic the statement $P implies Q$ will be true if $P$ and $Q$ are both true, or if $P$ is false. It will only be false if $P$ is true and $Q$ is false. If $P= : x in emptyset$ and $Q = : x le r :forall r in mathbb R$ then $P$ is always false. And $Q$ can never be true for any actuall $x in mathbb R$. So $P implies Q$ is true.
Also a statement is equivalent to the contrapostive. The contrapositive of $xin emptyset implies x le r:forall r in mathbb R$ is:
$exists rin mathbb R: r < x implies xnot in emptyset$. Well that is certainly true! If $r < x$ then $x$ must exist. And if $x$ exists then $x not in emptyset$!.
......
If $xin emptyset$ then $x$ is a green cheese eating alien who is the reincarnation of Elvis Presley is true because logically a false premise implies all conclusions.
So if $x in emptyset$ then $x le r$ for all $r in mathbb R$. (it's also true that $x >r$ fora all $r in mathbb R$ and that $x = r$ for all $r in mathbb R$ and so on.... as $x$ does not exist we can say anything we want to be true about it.)
....
Or looking at it another way: If $A_r = (-infty, r)$ then $emptyset subset A_r$ because the empty set is a subset of all sets. (The emptyset has no elements, so no elements aren't in any set.) So if $x in emptyset implies x in A_r$.
That is true for all $A_r: rin mathbb R$. So $x in (-infty, r)$ for all $rin mathbb R$ so $x le r$ for all $r in mathbb R$.
You might say "there's got to be a trick in there somewhere; nothing can be in all of those intervals" and you'd be right. The trick is that no such $xin emptyset$ does exist. But if such an $x$ DID exist, it would have to be in every set including every interval including those intervals.
answered Jan 24 at 17:07
fleabloodfleablood
72k22687
$endgroup$
$begingroup$
Why is it logical to say : a false premise implies all conclusions?
$endgroup$
– New2Math
Jan 24 at 19:59
$begingroup$
because it is..... $P$ implies $Q$ can be proven two ways, i) by checking that every time $P$ is true so is $Q$. We can't perform that test because $P$ is never true. or ii) by checking there's never a time when $P$ is true but $Q$ is false. And it's obvious there is never a time when $P$ is true but $Q$ is false because there is never a time that $P$ is true.
$endgroup$
– fleablood
Jan 24 at 20:08
add a comment |
$begingroup$
Three explanations: (But they all involve that as there are no $x in emptyset$ we can not negate $x le r$ for any $r in mathbb R$)
....
Either $xin emptyset implies x le r :forall r in mathbb R$ is true or it is false.
If it is false then that implies that there is an $x in emptyset$ and an $r in mathbb R$ so that $x > r$. That is impossible as there are no $x in emptyset$.
If it is true then for every $x in emptyset$ then ... something. We can never test this because we can never find an $x in emptyset$.
However we can do logic the statement $P implies Q$ will be true if $P$ and $Q$ are both true, or if $P$ is false. It will only be false if $P$ is true and $Q$ is false. If $P= : x in emptyset$ and $Q = : x le r :forall r in mathbb R$ then $P$ is always false. And $Q$ can never be true for any actuall $x in mathbb R$. So $P implies Q$ is true.
Also a statement is equivalent to the contrapostive. The contrapositive of $xin emptyset implies x le r:forall r in mathbb R$ is:
$exists rin mathbb R: r < x implies xnot in emptyset$. Well that is certainly true! If $r < x$ then $x$ must exist. And if $x$ exists then $x not in emptyset$!.
......
If $xin emptyset$ then $x$ is a green cheese eating alien who is the reincarnation of Elvis Presley is true because logically a false premise implies all conclusions.
So if $x in emptyset$ then $x le r$ for all $r in mathbb R$. (it's also true that $x >r$ fora all $r in mathbb R$ and that $x = r$ for all $r in mathbb R$ and so on.... as $x$ does not exist we can say anything we want to be true about it.)
....
Or looking at it another way: If $A_r = (-infty, r)$ then $emptyset subset A_r$ because the empty set is a subset of all sets. (The emptyset has no elements, so no elements aren't in any set.) So if $x in emptyset implies x in A_r$.
That is true for all $A_r: rin mathbb R$. So $x in (-infty, r)$ for all $rin mathbb R$ so $x le r$ for all $r in mathbb R$.
You might say "there's got to be a trick in there somewhere; nothing can be in all of those intervals" and you'd be right. The trick is that no such $xin emptyset$ does exist. But if such an $x$ DID exist, it would have to be in every set including every interval including those intervals.
answered Jan 24 at 17:07
fleabloodfleablood
72k22687
$endgroup$
Three explanations: (But they all involve that as there are no $x in emptyset$ we can not negate $x le r$ for any $r in mathbb R$)
....
Either $xin emptyset implies x le r :forall r in mathbb R$ is true or it is false.
If it is false then that implies that there is an $x in emptyset$ and an $r in mathbb R$ so that $x > r$. That is impossible as there are no $x in emptyset$.
If it is true then for every $x in emptyset$ then ... something. We can never test this because we can never find an $x in emptyset$.
However we can do logic the statement $P implies Q$ will be true if $P$ and $Q$ are both true, or if $P$ is false. It will only be false if $P$ is true and $Q$ is false. If $P= : x in emptyset$ and $Q = : x le r :forall r in mathbb R$ then $P$ is always false. And $Q$ can never be true for any actuall $x in mathbb R$. So $P implies Q$ is true.
Also a statement is equivalent to the contrapostive. The contrapositive of $xin emptyset implies x le r:forall r in mathbb R$ is:
$exists rin mathbb R: r < x implies xnot in emptyset$. Well that is certainly true! If $r < x$ then $x$ must exist. And if $x$ exists then $x not in emptyset$!.
......
If $xin emptyset$ then $x$ is a green cheese eating alien who is the reincarnation of Elvis Presley is true because logically a false premise implies all conclusions.
So if $x in emptyset$ then $x le r$ for all $r in mathbb R$. (it's also true that $x >r$ fora all $r in mathbb R$ and that $x = r$ for all $r in mathbb R$ and so on.... as $x$ does not exist we can say anything we want to be true about it.)
....
Or looking at it another way: If $A_r = (-infty, r)$ then $emptyset subset A_r$ because the empty set is a subset of all sets. (The emptyset has no elements, so no elements aren't in any set.) So if $x in emptyset implies x in A_r$.
That is true for all $A_r: rin mathbb R$. So $x in (-infty, r)$ for all $rin mathbb R$ so $x le r$ for all $r in mathbb R$.
You might say "there's got to be a trick in there somewhere; nothing can be in all of those intervals" and you'd be right. The trick is that no such $xin emptyset$ does exist. But if such an $x$ DID exist, it would have to be in every set including every interval including those intervals.
answered Jan 24 at 17:07
fleabloodfleablood
72k22687
edited Jan 24 at 17:29
answered Jan 24 at 17:07
fleabloodfleablood
72k22687
answered Jan 24 at 17:07
fleabloodfleablood
72k22687
answered Jan 24 at 17:07
fleabloodfleablood
72k22687
72k22687
$begingroup$
Why is it logical to say : a false premise implies all conclusions?
$endgroup$
– New2Math
Jan 24 at 19:59
$begingroup$
because it is..... $P$ implies $Q$ can be proven two ways, i) by checking that every time $P$ is true so is $Q$. We can't perform that test because $P$ is never true. or ii) by checking there's never a time when $P$ is true but $Q$ is false. And it's obvious there is never a time when $P$ is true but $Q$ is false because there is never a time that $P$ is true.
$endgroup$
– fleablood
Jan 24 at 20:08
add a comment |
$begingroup$
Why is it logical to say : a false premise implies all conclusions?
$endgroup$
– New2Math
Jan 24 at 19:59
$begingroup$
because it is..... $P$ implies $Q$ can be proven two ways, i) by checking that every time $P$ is true so is $Q$. We can't perform that test because $P$ is never true. or ii) by checking there's never a time when $P$ is true but $Q$ is false. And it's obvious there is never a time when $P$ is true but $Q$ is false because there is never a time that $P$ is true.
$endgroup$
– fleablood
Jan 24 at 20:08
$begingroup$
Why is it logical to say : a false premise implies all conclusions?
$endgroup$
– New2Math
Jan 24 at 19:59
$begingroup$
Why is it logical to say : a false premise implies all conclusions?
$endgroup$
– New2Math
Jan 24 at 19:59
$begingroup$
because it is..... $P$ implies $Q$ can be proven two ways, i) by checking that every time $P$ is true so is $Q$. We can't perform that test because $P$ is never true. or ii) by checking there's never a time when $P$ is true but $Q$ is false. And it's obvious there is never a time when $P$ is true but $Q$ is false because there is never a time that $P$ is true.
$endgroup$
– fleablood
Jan 24 at 20:08
$begingroup$
because it is..... $P$ implies $Q$ can be proven two ways, i) by checking that every time $P$ is true so is $Q$. We can't perform that test because $P$ is never true. or ii) by checking there's never a time when $P$ is true but $Q$ is false. And it's obvious there is never a time when $P$ is true but $Q$ is false because there is never a time that $P$ is true.
$endgroup$
– fleablood
Jan 24 at 20:08
add a comment |
$begingroup$
Given $r$, every element of the empty set is at most $r$. This statement is true because its negation is false. Namely, its negation is that there exists $r$ and an element of the empty set that is at greater than $r$. But the empty set has no elements.
answered Jan 24 at 16:39
Foobaz JohnFoobaz John
22.5k41452
$endgroup$
add a comment |
$begingroup$
Given $r$, every element of the empty set is at most $r$. This statement is true because its negation is false. Namely, its negation is that there exists $r$ and an element of the empty set that is at greater than $r$. But the empty set has no elements.
answered Jan 24 at 16:39
Foobaz JohnFoobaz John
22.5k41452
$endgroup$
add a comment |
$begingroup$
Given $r$, every element of the empty set is at most $r$. This statement is true because its negation is false. Namely, its negation is that there exists $r$ and an element of the empty set that is at greater than $r$. But the empty set has no elements.
answered Jan 24 at 16:39
Foobaz JohnFoobaz John
22.5k41452
$endgroup$
Given $r$, every element of the empty set is at most $r$. This statement is true because its negation is false. Namely, its negation is that there exists $r$ and an element of the empty set that is at greater than $r$. But the empty set has no elements.
answered Jan 24 at 16:39
Foobaz JohnFoobaz John
22.5k41452
answered Jan 24 at 16:39
Foobaz JohnFoobaz John
22.5k41452
answered Jan 24 at 16:39
Foobaz JohnFoobaz John
22.5k41452
answered Jan 24 at 16:39
Foobaz JohnFoobaz John
22.5k41452
22.5k41452
add a comment |
add a comment |
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1
$begingroup$
You have to consider the condition : "if $x in emptyset$, then $x le r$", for an $r in mathbb R$ whatever.
$endgroup$
– Mauro ALLEGRANZA
Jan 24 at 16:33
1
$begingroup$
Due to the fact that $x in emptyset$ is always false the above condition is always true, for every $r in mathbb R$, whatever small $r$ is.
$endgroup$
– Mauro ALLEGRANZA
Jan 24 at 16:34