Mixing
Proof that $inf A = -sup(-A)$
$begingroup$
Let $A$ be a nonempty subset of real numbers which is bounded below. Let $-A$ be the set of of all numbers $-x$, where $x$ is in $A$. Prove that $inf A = -sup(-A)$
So far this is what I have
Let $alpha=inf(A)$, which allows us to say that $alpha leq x$ for all $x in A$. Therefore, we know that $-alpha geq -x$ for all $x in -A$. Therefore we know that $-alpha$ is an upper bound of $-A$. $ $
Now let $b$ be the upper bound of $-A$. There exists $b geq-x implies-b leq x$ for all $x in A$. Hence,
begin{align}
-b & leq alpha\
-alpha & leq b\
-alpha & = - inf(A) = sup (-A)
end{align}
By multiplying $-1$ on both sides, we get that $inf(A) = -sup (-A)$
Is my proof correct?
real-analysis proof-verification supremum-and-infimum
edited Jan 16 '17 at 14:14
Martin Sleziak
44.9k10122277
asked May 15 '13 at 5:26
user77107user77107
279237
$endgroup$
add a comment |
$begingroup$
Let $A$ be a nonempty subset of real numbers which is bounded below. Let $-A$ be the set of of all numbers $-x$, where $x$ is in $A$. Prove that $inf A = -sup(-A)$
So far this is what I have
Let $alpha=inf(A)$, which allows us to say that $alpha leq x$ for all $x in A$. Therefore, we know that $-alpha geq -x$ for all $x in -A$. Therefore we know that $-alpha$ is an upper bound of $-A$. $ $
Now let $b$ be the upper bound of $-A$. There exists $b geq-x implies-b leq x$ for all $x in A$. Hence,
begin{align}
-b & leq alpha\
-alpha & leq b\
-alpha & = - inf(A) = sup (-A)
end{align}
By multiplying $-1$ on both sides, we get that $inf(A) = -sup (-A)$
Is my proof correct?
real-analysis proof-verification supremum-and-infimum
edited Jan 16 '17 at 14:14
Martin Sleziak
44.9k10122277
asked May 15 '13 at 5:26
user77107user77107
279237
$endgroup$
3
$begingroup$
+1. For showing your work. People appreciate if you show your work. And your answer is right. Good work!
$endgroup$
– user17762
May 15 '13 at 5:31
1
$begingroup$
how did you get $-b leq alpha$?
$endgroup$
– Real Hilbert
Nov 2 '13 at 15:41
$begingroup$
See math.stackexchange.com/a/947049/589.
$endgroup$
– lhf
Oct 31 '14 at 17:54
$begingroup$
"Therefore, we know that $-alpha geq −x$ for all $x in -A$." Consider $A = {1, 3, 5}$ then $alpha = mathrm{inf} A = 1$ and $-A = {-1, -3, -5}$ and you get a negation to much. With these variables $-1 geq −x$ for all $x in {-1, -3, -5}$ is not true.
$endgroup$
– Björn Lindqvist
Jun 19 '18 at 0:09
add a comment |
$begingroup$
Let $A$ be a nonempty subset of real numbers which is bounded below. Let $-A$ be the set of of all numbers $-x$, where $x$ is in $A$. Prove that $inf A = -sup(-A)$
So far this is what I have
Let $alpha=inf(A)$, which allows us to say that $alpha leq x$ for all $x in A$. Therefore, we know that $-alpha geq -x$ for all $x in -A$. Therefore we know that $-alpha$ is an upper bound of $-A$. $ $
Now let $b$ be the upper bound of $-A$. There exists $b geq-x implies-b leq x$ for all $x in A$. Hence,
begin{align}
-b & leq alpha\
-alpha & leq b\
-alpha & = - inf(A) = sup (-A)
end{align}
By multiplying $-1$ on both sides, we get that $inf(A) = -sup (-A)$
Is my proof correct?
real-analysis proof-verification supremum-and-infimum
edited Jan 16 '17 at 14:14
Martin Sleziak
44.9k10122277
asked May 15 '13 at 5:26
user77107user77107
279237
$endgroup$
Let $A$ be a nonempty subset of real numbers which is bounded below. Let $-A$ be the set of of all numbers $-x$, where $x$ is in $A$. Prove that $inf A = -sup(-A)$
So far this is what I have
Let $alpha=inf(A)$, which allows us to say that $alpha leq x$ for all $x in A$. Therefore, we know that $-alpha geq -x$ for all $x in -A$. Therefore we know that $-alpha$ is an upper bound of $-A$. $ $
Now let $b$ be the upper bound of $-A$. There exists $b geq-x implies-b leq x$ for all $x in A$. Hence,
begin{align}
-b & leq alpha\
-alpha & leq b\
-alpha & = - inf(A) = sup (-A)
end{align}
By multiplying $-1$ on both sides, we get that $inf(A) = -sup (-A)$
Is my proof correct?
real-analysis proof-verification supremum-and-infimum
real-analysis proof-verification supremum-and-infimum
edited Jan 16 '17 at 14:14
Martin Sleziak
44.9k10122277
asked May 15 '13 at 5:26
user77107user77107
279237
edited Jan 16 '17 at 14:14
Martin Sleziak
44.9k10122277
asked May 15 '13 at 5:26
user77107user77107
279237
edited Jan 16 '17 at 14:14
Martin Sleziak
44.9k10122277
edited Jan 16 '17 at 14:14
Martin Sleziak
44.9k10122277
edited Jan 16 '17 at 14:14
Martin Sleziak
44.9k10122277
44.9k10122277
asked May 15 '13 at 5:26
user77107user77107
279237
asked May 15 '13 at 5:26
user77107user77107
279237
asked May 15 '13 at 5:26
user77107user77107
279237
279237
3
$begingroup$
+1. For showing your work. People appreciate if you show your work. And your answer is right. Good work!
$endgroup$
– user17762
May 15 '13 at 5:31
1
$begingroup$
how did you get $-b leq alpha$?
$endgroup$
– Real Hilbert
Nov 2 '13 at 15:41
$begingroup$
See math.stackexchange.com/a/947049/589.
$endgroup$
– lhf
Oct 31 '14 at 17:54
$begingroup$
"Therefore, we know that $-alpha geq −x$ for all $x in -A$." Consider $A = {1, 3, 5}$ then $alpha = mathrm{inf} A = 1$ and $-A = {-1, -3, -5}$ and you get a negation to much. With these variables $-1 geq −x$ for all $x in {-1, -3, -5}$ is not true.
$endgroup$
– Björn Lindqvist
Jun 19 '18 at 0:09
add a comment |
3
$begingroup$
+1. For showing your work. People appreciate if you show your work. And your answer is right. Good work!
$endgroup$
– user17762
May 15 '13 at 5:31
1
$begingroup$
how did you get $-b leq alpha$?
$endgroup$
– Real Hilbert
Nov 2 '13 at 15:41
$begingroup$
See math.stackexchange.com/a/947049/589.
$endgroup$
– lhf
Oct 31 '14 at 17:54
$begingroup$
"Therefore, we know that $-alpha geq −x$ for all $x in -A$." Consider $A = {1, 3, 5}$ then $alpha = mathrm{inf} A = 1$ and $-A = {-1, -3, -5}$ and you get a negation to much. With these variables $-1 geq −x$ for all $x in {-1, -3, -5}$ is not true.
$endgroup$
– Björn Lindqvist
Jun 19 '18 at 0:09
3
3
$begingroup$
+1. For showing your work. People appreciate if you show your work. And your answer is right. Good work!
$endgroup$
– user17762
May 15 '13 at 5:31
$begingroup$
+1. For showing your work. People appreciate if you show your work. And your answer is right. Good work!
$endgroup$
– user17762
May 15 '13 at 5:31
1
1
$begingroup$
how did you get $-b leq alpha$?
$endgroup$
– Real Hilbert
Nov 2 '13 at 15:41
$begingroup$
how did you get $-b leq alpha$?
$endgroup$
– Real Hilbert
Nov 2 '13 at 15:41
$begingroup$
See math.stackexchange.com/a/947049/589.
$endgroup$
– lhf
Oct 31 '14 at 17:54
$begingroup$
See math.stackexchange.com/a/947049/589.
$endgroup$
– lhf
Oct 31 '14 at 17:54
$begingroup$
"Therefore, we know that $-alpha geq −x$ for all $x in -A$." Consider $A = {1, 3, 5}$ then $alpha = mathrm{inf} A = 1$ and $-A = {-1, -3, -5}$ and you get a negation to much. With these variables $-1 geq −x$ for all $x in {-1, -3, -5}$ is not true.
$endgroup$
– Björn Lindqvist
Jun 19 '18 at 0:09
$begingroup$
"Therefore, we know that $-alpha geq −x$ for all $x in -A$." Consider $A = {1, 3, 5}$ then $alpha = mathrm{inf} A = 1$ and $-A = {-1, -3, -5}$ and you get a negation to much. With these variables $-1 geq −x$ for all $x in {-1, -3, -5}$ is not true.
$endgroup$
– Björn Lindqvist
Jun 19 '18 at 0:09
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Let $A$ be a nonempty subset of real numbers which is bounded below. Let $−A$ be the set of of all numbers $−x$, where $x$ is in $A$. Prove that $inf{A}=−sup{(-A)}$.
I think that the purpose of this question is to show you why it is not required to include the existence of infimum into the Axiom of Completeness.
Here, i will show the existence of infimum of $A$.
$forall ain A, exists xinmathbb{R}text{ such that } xleq a implies -xgeq-a implies -x text{ is an upper bound for }-A.$
Here, $x$ is an arbitrary lower bound for $A$.
By axiom of completeness, $exists yin mathbb{R} text{ such that }y=sup{(-A)}, text{i.e. }-aleq yleq -x text{ , }forall ain A,$
which implies that $xleq -y leq a$.
Here, $-y$ is a lower bound for $A$ and $-y$ is at least any lower bound for $A$, which means that $-y = inf{A}$.
Hence, $-sup{(-A)} =inf{A}$.
answered Jan 31 '17 at 10:28
Little RookieLittle Rookie
617724
$endgroup$
add a comment |
$begingroup$
Aside from dropping a $-$ sign in your last line, that's good work. (Also, I'd say "let $b$ be an upper bound," instead, but it's clear that's what is meant.)
$endgroup$
add a comment |
$begingroup$
Not a direct answer to your question, but here is essentially your proof written down in an alternative style, which might be helpful. (See, e.g., EWD1300 for more details on this proof style.)$
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{ref}[1]{text{(#1)}}
newcommand{inf}[1]{text{inf}(#1)}
newcommand{sup}[1]{text{sup}(#1)}
$
First, let's make our definitions explicit. The simplest definitions of $;inf{cdot};$ and $;sup{cdot};$ I know are
begin{align}
tag{0} z leq inf{A} ;equiv; langle forall a : a in A : z leq a rangle
\
tag{1} sup{B} leq z ;equiv; langle forall b : b in B : b leq z rangle
end{align}
for any real $;z;$, lower-bounded non-empty $;A;$ and upper-bounded non-empty $;B;$.
Armed with these definitions, let's try to lower-bound both sides, starting with the most complex side, so the right hand side: for any $;z;$,
$$calc
z leq -sup{-A}
op=hints{negate both sides and swap them}hint{-- preparing for the definition of $;sup{cdot};$}
sup{-A} leq -z
op=hints{definition $ref{1}$}hints{-- allowed because $;-A;$ is upper-bounded and non-empty,}hints{because $;A;$ is lower-bounded and non-empty}hint{by the definition of $;-A;$}
langle forall b : b in -A : b leq -z rangle
op=hint{definition of $;-A;$}
langle forall b : -b in A : b leq -z rangle
op=hint{substitute $;a := -b;$; simplify using arithmetic}
langle forall a : a in A : z leq a rangle
op=hints{definition $ref{0}$}hint{-- allowed because $;A;$ is lower-bounded and non-empty}
z leq inf{A}
endcalc$$
In other words, $;-sup{-A};$ and $;inf{A};$ have the same lower bounds, and therefore they are equal.
This proof uses the following principle:
$$
x = y ;equiv; langle forall z :: z leq x ;equiv; z leq y rangle
$$
answered Apr 28 '16 at 9:41
Marnix KloosterMarnix Klooster
4,25122149
$endgroup$
add a comment |
$begingroup$
your method was good but you didn't check it . i try to proof it well :
( sorry my english is not well because i am not a english person )
$a=inf(A)$ so $ale x$ for all $xin A$. Therefore, $−ale −x$ for all $-xin -A$.
Now let $b$ be the upper bound of $−A$. There exists $bge−x$ $−ble x$ for all $xin .
A$. Hence,
$−b−a−a≤a≤b=−inf(A)=sup(−A)$
By multiplying $−1$ on both sides, we get that $inf(A)=−sup(−A)$
edited Apr 28 '16 at 10:20
InsideOut
5,11431034
answered Oct 31 '14 at 17:36
amirhossein nasiriamirhossein nasiri
111
$endgroup$
add a comment |
$begingroup$
Correct me if wrong:
$Asubset mathbb{R}$, and $cA=${$y|y=cx, x in A$}.
Let $c <0$.
Show that $sup (cA)= c inf (A)$.
Then
$cx le M$ $iff$ $x ge M/c$, i.e.
$M$ is an upper bound of $cA$
$ iff$
$M/c$ is a lower bound of $A$.
1)Let $M= sup (cA)$, then
$x ge sup (cA)/c$, $x in A$.
Hence $inf A ge sup (cA)$.
(Definition of $inf A$).
2) Let $M/c =inf A$, then
$cx le M = c inf A$.
Hence $sup (cA) le c inf A$ .
(Definition of $sup$)
Finally 1) and 2):
$sup (cA)=c inf A$.
($c=-1$ in the original problem)
answered Jan 30 at 10:46
Peter SzilasPeter Szilas
11.7k2822
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $A$ be a nonempty subset of real numbers which is bounded below. Let $−A$ be the set of of all numbers $−x$, where $x$ is in $A$. Prove that $inf{A}=−sup{(-A)}$.
I think that the purpose of this question is to show you why it is not required to include the existence of infimum into the Axiom of Completeness.
Here, i will show the existence of infimum of $A$.
$forall ain A, exists xinmathbb{R}text{ such that } xleq a implies -xgeq-a implies -x text{ is an upper bound for }-A.$
Here, $x$ is an arbitrary lower bound for $A$.
By axiom of completeness, $exists yin mathbb{R} text{ such that }y=sup{(-A)}, text{i.e. }-aleq yleq -x text{ , }forall ain A,$
which implies that $xleq -y leq a$.
Here, $-y$ is a lower bound for $A$ and $-y$ is at least any lower bound for $A$, which means that $-y = inf{A}$.
Hence, $-sup{(-A)} =inf{A}$.
answered Jan 31 '17 at 10:28
Little RookieLittle Rookie
617724
$endgroup$
add a comment |
$begingroup$
Let $A$ be a nonempty subset of real numbers which is bounded below. Let $−A$ be the set of of all numbers $−x$, where $x$ is in $A$. Prove that $inf{A}=−sup{(-A)}$.
I think that the purpose of this question is to show you why it is not required to include the existence of infimum into the Axiom of Completeness.
Here, i will show the existence of infimum of $A$.
$forall ain A, exists xinmathbb{R}text{ such that } xleq a implies -xgeq-a implies -x text{ is an upper bound for }-A.$
Here, $x$ is an arbitrary lower bound for $A$.
By axiom of completeness, $exists yin mathbb{R} text{ such that }y=sup{(-A)}, text{i.e. }-aleq yleq -x text{ , }forall ain A,$
which implies that $xleq -y leq a$.
Here, $-y$ is a lower bound for $A$ and $-y$ is at least any lower bound for $A$, which means that $-y = inf{A}$.
Hence, $-sup{(-A)} =inf{A}$.
answered Jan 31 '17 at 10:28
Little RookieLittle Rookie
617724
$endgroup$
add a comment |
$begingroup$
Let $A$ be a nonempty subset of real numbers which is bounded below. Let $−A$ be the set of of all numbers $−x$, where $x$ is in $A$. Prove that $inf{A}=−sup{(-A)}$.
I think that the purpose of this question is to show you why it is not required to include the existence of infimum into the Axiom of Completeness.
Here, i will show the existence of infimum of $A$.
$forall ain A, exists xinmathbb{R}text{ such that } xleq a implies -xgeq-a implies -x text{ is an upper bound for }-A.$
Here, $x$ is an arbitrary lower bound for $A$.
By axiom of completeness, $exists yin mathbb{R} text{ such that }y=sup{(-A)}, text{i.e. }-aleq yleq -x text{ , }forall ain A,$
which implies that $xleq -y leq a$.
Here, $-y$ is a lower bound for $A$ and $-y$ is at least any lower bound for $A$, which means that $-y = inf{A}$.
Hence, $-sup{(-A)} =inf{A}$.
answered Jan 31 '17 at 10:28
Little RookieLittle Rookie
617724
$endgroup$
Let $A$ be a nonempty subset of real numbers which is bounded below. Let $−A$ be the set of of all numbers $−x$, where $x$ is in $A$. Prove that $inf{A}=−sup{(-A)}$.
I think that the purpose of this question is to show you why it is not required to include the existence of infimum into the Axiom of Completeness.
Here, i will show the existence of infimum of $A$.
$forall ain A, exists xinmathbb{R}text{ such that } xleq a implies -xgeq-a implies -x text{ is an upper bound for }-A.$
Here, $x$ is an arbitrary lower bound for $A$.
By axiom of completeness, $exists yin mathbb{R} text{ such that }y=sup{(-A)}, text{i.e. }-aleq yleq -x text{ , }forall ain A,$
which implies that $xleq -y leq a$.
Here, $-y$ is a lower bound for $A$ and $-y$ is at least any lower bound for $A$, which means that $-y = inf{A}$.
Hence, $-sup{(-A)} =inf{A}$.
answered Jan 31 '17 at 10:28
Little RookieLittle Rookie
617724
edited Jan 31 '17 at 11:42
answered Jan 31 '17 at 10:28
Little RookieLittle Rookie
617724
answered Jan 31 '17 at 10:28
Little RookieLittle Rookie
617724
answered Jan 31 '17 at 10:28
Little RookieLittle Rookie
617724
617724
add a comment |
add a comment |
$begingroup$
Aside from dropping a $-$ sign in your last line, that's good work. (Also, I'd say "let $b$ be an upper bound," instead, but it's clear that's what is meant.)
$endgroup$
add a comment |
$begingroup$
Aside from dropping a $-$ sign in your last line, that's good work. (Also, I'd say "let $b$ be an upper bound," instead, but it's clear that's what is meant.)
$endgroup$
add a comment |
$begingroup$
Aside from dropping a $-$ sign in your last line, that's good work. (Also, I'd say "let $b$ be an upper bound," instead, but it's clear that's what is meant.)
$endgroup$
Aside from dropping a $-$ sign in your last line, that's good work. (Also, I'd say "let $b$ be an upper bound," instead, but it's clear that's what is meant.)
answered May 15 '13 at 5:28
community wiki
Cameron Buie
add a comment |
add a comment |
$begingroup$
Not a direct answer to your question, but here is essentially your proof written down in an alternative style, which might be helpful. (See, e.g., EWD1300 for more details on this proof style.)$
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{ref}[1]{text{(#1)}}
newcommand{inf}[1]{text{inf}(#1)}
newcommand{sup}[1]{text{sup}(#1)}
$
First, let's make our definitions explicit. The simplest definitions of $;inf{cdot};$ and $;sup{cdot};$ I know are
begin{align}
tag{0} z leq inf{A} ;equiv; langle forall a : a in A : z leq a rangle
\
tag{1} sup{B} leq z ;equiv; langle forall b : b in B : b leq z rangle
end{align}
for any real $;z;$, lower-bounded non-empty $;A;$ and upper-bounded non-empty $;B;$.
Armed with these definitions, let's try to lower-bound both sides, starting with the most complex side, so the right hand side: for any $;z;$,
$$calc
z leq -sup{-A}
op=hints{negate both sides and swap them}hint{-- preparing for the definition of $;sup{cdot};$}
sup{-A} leq -z
op=hints{definition $ref{1}$}hints{-- allowed because $;-A;$ is upper-bounded and non-empty,}hints{because $;A;$ is lower-bounded and non-empty}hint{by the definition of $;-A;$}
langle forall b : b in -A : b leq -z rangle
op=hint{definition of $;-A;$}
langle forall b : -b in A : b leq -z rangle
op=hint{substitute $;a := -b;$; simplify using arithmetic}
langle forall a : a in A : z leq a rangle
op=hints{definition $ref{0}$}hint{-- allowed because $;A;$ is lower-bounded and non-empty}
z leq inf{A}
endcalc$$
In other words, $;-sup{-A};$ and $;inf{A};$ have the same lower bounds, and therefore they are equal.
This proof uses the following principle:
$$
x = y ;equiv; langle forall z :: z leq x ;equiv; z leq y rangle
$$
answered Apr 28 '16 at 9:41
Marnix KloosterMarnix Klooster
4,25122149
$endgroup$
add a comment |
$begingroup$
Not a direct answer to your question, but here is essentially your proof written down in an alternative style, which might be helpful. (See, e.g., EWD1300 for more details on this proof style.)$
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{ref}[1]{text{(#1)}}
newcommand{inf}[1]{text{inf}(#1)}
newcommand{sup}[1]{text{sup}(#1)}
$
First, let's make our definitions explicit. The simplest definitions of $;inf{cdot};$ and $;sup{cdot};$ I know are
begin{align}
tag{0} z leq inf{A} ;equiv; langle forall a : a in A : z leq a rangle
\
tag{1} sup{B} leq z ;equiv; langle forall b : b in B : b leq z rangle
end{align}
for any real $;z;$, lower-bounded non-empty $;A;$ and upper-bounded non-empty $;B;$.
Armed with these definitions, let's try to lower-bound both sides, starting with the most complex side, so the right hand side: for any $;z;$,
$$calc
z leq -sup{-A}
op=hints{negate both sides and swap them}hint{-- preparing for the definition of $;sup{cdot};$}
sup{-A} leq -z
op=hints{definition $ref{1}$}hints{-- allowed because $;-A;$ is upper-bounded and non-empty,}hints{because $;A;$ is lower-bounded and non-empty}hint{by the definition of $;-A;$}
langle forall b : b in -A : b leq -z rangle
op=hint{definition of $;-A;$}
langle forall b : -b in A : b leq -z rangle
op=hint{substitute $;a := -b;$; simplify using arithmetic}
langle forall a : a in A : z leq a rangle
op=hints{definition $ref{0}$}hint{-- allowed because $;A;$ is lower-bounded and non-empty}
z leq inf{A}
endcalc$$
In other words, $;-sup{-A};$ and $;inf{A};$ have the same lower bounds, and therefore they are equal.
This proof uses the following principle:
$$
x = y ;equiv; langle forall z :: z leq x ;equiv; z leq y rangle
$$
answered Apr 28 '16 at 9:41
Marnix KloosterMarnix Klooster
4,25122149
$endgroup$
add a comment |
$begingroup$
Not a direct answer to your question, but here is essentially your proof written down in an alternative style, which might be helpful. (See, e.g., EWD1300 for more details on this proof style.)$
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{ref}[1]{text{(#1)}}
newcommand{inf}[1]{text{inf}(#1)}
newcommand{sup}[1]{text{sup}(#1)}
$
First, let's make our definitions explicit. The simplest definitions of $;inf{cdot};$ and $;sup{cdot};$ I know are
begin{align}
tag{0} z leq inf{A} ;equiv; langle forall a : a in A : z leq a rangle
\
tag{1} sup{B} leq z ;equiv; langle forall b : b in B : b leq z rangle
end{align}
for any real $;z;$, lower-bounded non-empty $;A;$ and upper-bounded non-empty $;B;$.
Armed with these definitions, let's try to lower-bound both sides, starting with the most complex side, so the right hand side: for any $;z;$,
$$calc
z leq -sup{-A}
op=hints{negate both sides and swap them}hint{-- preparing for the definition of $;sup{cdot};$}
sup{-A} leq -z
op=hints{definition $ref{1}$}hints{-- allowed because $;-A;$ is upper-bounded and non-empty,}hints{because $;A;$ is lower-bounded and non-empty}hint{by the definition of $;-A;$}
langle forall b : b in -A : b leq -z rangle
op=hint{definition of $;-A;$}
langle forall b : -b in A : b leq -z rangle
op=hint{substitute $;a := -b;$; simplify using arithmetic}
langle forall a : a in A : z leq a rangle
op=hints{definition $ref{0}$}hint{-- allowed because $;A;$ is lower-bounded and non-empty}
z leq inf{A}
endcalc$$
In other words, $;-sup{-A};$ and $;inf{A};$ have the same lower bounds, and therefore they are equal.
This proof uses the following principle:
$$
x = y ;equiv; langle forall z :: z leq x ;equiv; z leq y rangle
$$
answered Apr 28 '16 at 9:41
Marnix KloosterMarnix Klooster
4,25122149
$endgroup$
Not a direct answer to your question, but here is essentially your proof written down in an alternative style, which might be helpful. (See, e.g., EWD1300 for more details on this proof style.)$
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{ref}[1]{text{(#1)}}
newcommand{inf}[1]{text{inf}(#1)}
newcommand{sup}[1]{text{sup}(#1)}
$
First, let's make our definitions explicit. The simplest definitions of $;inf{cdot};$ and $;sup{cdot};$ I know are
begin{align}
tag{0} z leq inf{A} ;equiv; langle forall a : a in A : z leq a rangle
\
tag{1} sup{B} leq z ;equiv; langle forall b : b in B : b leq z rangle
end{align}
for any real $;z;$, lower-bounded non-empty $;A;$ and upper-bounded non-empty $;B;$.
Armed with these definitions, let's try to lower-bound both sides, starting with the most complex side, so the right hand side: for any $;z;$,
$$calc
z leq -sup{-A}
op=hints{negate both sides and swap them}hint{-- preparing for the definition of $;sup{cdot};$}
sup{-A} leq -z
op=hints{definition $ref{1}$}hints{-- allowed because $;-A;$ is upper-bounded and non-empty,}hints{because $;A;$ is lower-bounded and non-empty}hint{by the definition of $;-A;$}
langle forall b : b in -A : b leq -z rangle
op=hint{definition of $;-A;$}
langle forall b : -b in A : b leq -z rangle
op=hint{substitute $;a := -b;$; simplify using arithmetic}
langle forall a : a in A : z leq a rangle
op=hints{definition $ref{0}$}hint{-- allowed because $;A;$ is lower-bounded and non-empty}
z leq inf{A}
endcalc$$
In other words, $;-sup{-A};$ and $;inf{A};$ have the same lower bounds, and therefore they are equal.
This proof uses the following principle:
$$
x = y ;equiv; langle forall z :: z leq x ;equiv; z leq y rangle
$$
answered Apr 28 '16 at 9:41
Marnix KloosterMarnix Klooster
4,25122149
answered Apr 28 '16 at 9:41
Marnix KloosterMarnix Klooster
4,25122149
answered Apr 28 '16 at 9:41
Marnix KloosterMarnix Klooster
4,25122149
answered Apr 28 '16 at 9:41
Marnix KloosterMarnix Klooster
4,25122149
4,25122149
add a comment |
add a comment |
$begingroup$
your method was good but you didn't check it . i try to proof it well :
( sorry my english is not well because i am not a english person )
$a=inf(A)$ so $ale x$ for all $xin A$. Therefore, $−ale −x$ for all $-xin -A$.
Now let $b$ be the upper bound of $−A$. There exists $bge−x$ $−ble x$ for all $xin .
A$. Hence,
$−b−a−a≤a≤b=−inf(A)=sup(−A)$
By multiplying $−1$ on both sides, we get that $inf(A)=−sup(−A)$
edited Apr 28 '16 at 10:20
InsideOut
5,11431034
answered Oct 31 '14 at 17:36
amirhossein nasiriamirhossein nasiri
111
$endgroup$
add a comment |
$begingroup$
your method was good but you didn't check it . i try to proof it well :
( sorry my english is not well because i am not a english person )
$a=inf(A)$ so $ale x$ for all $xin A$. Therefore, $−ale −x$ for all $-xin -A$.
Now let $b$ be the upper bound of $−A$. There exists $bge−x$ $−ble x$ for all $xin .
A$. Hence,
$−b−a−a≤a≤b=−inf(A)=sup(−A)$
By multiplying $−1$ on both sides, we get that $inf(A)=−sup(−A)$
edited Apr 28 '16 at 10:20
InsideOut
5,11431034
answered Oct 31 '14 at 17:36
amirhossein nasiriamirhossein nasiri
111
$endgroup$
add a comment |
$begingroup$
your method was good but you didn't check it . i try to proof it well :
( sorry my english is not well because i am not a english person )
$a=inf(A)$ so $ale x$ for all $xin A$. Therefore, $−ale −x$ for all $-xin -A$.
Now let $b$ be the upper bound of $−A$. There exists $bge−x$ $−ble x$ for all $xin .
A$. Hence,
$−b−a−a≤a≤b=−inf(A)=sup(−A)$
By multiplying $−1$ on both sides, we get that $inf(A)=−sup(−A)$
edited Apr 28 '16 at 10:20
InsideOut
5,11431034
answered Oct 31 '14 at 17:36
amirhossein nasiriamirhossein nasiri
111
$endgroup$
your method was good but you didn't check it . i try to proof it well :
( sorry my english is not well because i am not a english person )
$a=inf(A)$ so $ale x$ for all $xin A$. Therefore, $−ale −x$ for all $-xin -A$.
Now let $b$ be the upper bound of $−A$. There exists $bge−x$ $−ble x$ for all $xin .
A$. Hence,
$−b−a−a≤a≤b=−inf(A)=sup(−A)$
By multiplying $−1$ on both sides, we get that $inf(A)=−sup(−A)$
edited Apr 28 '16 at 10:20
InsideOut
5,11431034
answered Oct 31 '14 at 17:36
amirhossein nasiriamirhossein nasiri
111
edited Apr 28 '16 at 10:20
InsideOut
5,11431034
edited Apr 28 '16 at 10:20
InsideOut
5,11431034
edited Apr 28 '16 at 10:20
InsideOut
5,11431034
5,11431034
answered Oct 31 '14 at 17:36
amirhossein nasiriamirhossein nasiri
111
answered Oct 31 '14 at 17:36
amirhossein nasiriamirhossein nasiri
111
answered Oct 31 '14 at 17:36
amirhossein nasiriamirhossein nasiri
111
111
add a comment |
add a comment |
$begingroup$
Correct me if wrong:
$Asubset mathbb{R}$, and $cA=${$y|y=cx, x in A$}.
Let $c <0$.
Show that $sup (cA)= c inf (A)$.
Then
$cx le M$ $iff$ $x ge M/c$, i.e.
$M$ is an upper bound of $cA$
$ iff$
$M/c$ is a lower bound of $A$.
1)Let $M= sup (cA)$, then
$x ge sup (cA)/c$, $x in A$.
Hence $inf A ge sup (cA)$.
(Definition of $inf A$).
2) Let $M/c =inf A$, then
$cx le M = c inf A$.
Hence $sup (cA) le c inf A$ .
(Definition of $sup$)
Finally 1) and 2):
$sup (cA)=c inf A$.
($c=-1$ in the original problem)
answered Jan 30 at 10:46
Peter SzilasPeter Szilas
11.7k2822
$endgroup$
add a comment |
$begingroup$
Correct me if wrong:
$Asubset mathbb{R}$, and $cA=${$y|y=cx, x in A$}.
Let $c <0$.
Show that $sup (cA)= c inf (A)$.
Then
$cx le M$ $iff$ $x ge M/c$, i.e.
$M$ is an upper bound of $cA$
$ iff$
$M/c$ is a lower bound of $A$.
1)Let $M= sup (cA)$, then
$x ge sup (cA)/c$, $x in A$.
Hence $inf A ge sup (cA)$.
(Definition of $inf A$).
2) Let $M/c =inf A$, then
$cx le M = c inf A$.
Hence $sup (cA) le c inf A$ .
(Definition of $sup$)
Finally 1) and 2):
$sup (cA)=c inf A$.
($c=-1$ in the original problem)
answered Jan 30 at 10:46
Peter SzilasPeter Szilas
11.7k2822
$endgroup$
add a comment |
$begingroup$
Correct me if wrong:
$Asubset mathbb{R}$, and $cA=${$y|y=cx, x in A$}.
Let $c <0$.
Show that $sup (cA)= c inf (A)$.
Then
$cx le M$ $iff$ $x ge M/c$, i.e.
$M$ is an upper bound of $cA$
$ iff$
$M/c$ is a lower bound of $A$.
1)Let $M= sup (cA)$, then
$x ge sup (cA)/c$, $x in A$.
Hence $inf A ge sup (cA)$.
(Definition of $inf A$).
2) Let $M/c =inf A$, then
$cx le M = c inf A$.
Hence $sup (cA) le c inf A$ .
(Definition of $sup$)
Finally 1) and 2):
$sup (cA)=c inf A$.
($c=-1$ in the original problem)
answered Jan 30 at 10:46
Peter SzilasPeter Szilas
11.7k2822
$endgroup$
Correct me if wrong:
$Asubset mathbb{R}$, and $cA=${$y|y=cx, x in A$}.
Let $c <0$.
Show that $sup (cA)= c inf (A)$.
Then
$cx le M$ $iff$ $x ge M/c$, i.e.
$M$ is an upper bound of $cA$
$ iff$
$M/c$ is a lower bound of $A$.
1)Let $M= sup (cA)$, then
$x ge sup (cA)/c$, $x in A$.
Hence $inf A ge sup (cA)$.
(Definition of $inf A$).
2) Let $M/c =inf A$, then
$cx le M = c inf A$.
Hence $sup (cA) le c inf A$ .
(Definition of $sup$)
Finally 1) and 2):
$sup (cA)=c inf A$.
($c=-1$ in the original problem)
answered Jan 30 at 10:46
Peter SzilasPeter Szilas
11.7k2822
answered Jan 30 at 10:46
Peter SzilasPeter Szilas
11.7k2822
answered Jan 30 at 10:46
Peter SzilasPeter Szilas
11.7k2822
answered Jan 30 at 10:46
Peter SzilasPeter Szilas
11.7k2822
11.7k2822
add a comment |
add a comment |
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$begingroup$
+1. For showing your work. People appreciate if you show your work. And your answer is right. Good work!
$endgroup$
– user17762
May 15 '13 at 5:31
1
$begingroup$
how did you get $-b leq alpha$?
$endgroup$
– Real Hilbert
Nov 2 '13 at 15:41
$begingroup$
See math.stackexchange.com/a/947049/589.
$endgroup$
– lhf
Oct 31 '14 at 17:54
$begingroup$
"Therefore, we know that $-alpha geq −x$ for all $x in -A$." Consider $A = {1, 3, 5}$ then $alpha = mathrm{inf} A = 1$ and $-A = {-1, -3, -5}$ and you get a negation to much. With these variables $-1 geq −x$ for all $x in {-1, -3, -5}$ is not true.
$endgroup$
– Björn Lindqvist
Jun 19 '18 at 0:09