Infimum and absolute values (Rudin's baby analysis)
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In the second last inequality of the proof of Theorem 6.17 (Rudin's baby analysis), I think he uses the fact that $$|x-y|leq c text{ implies } |inf x-inf y|leq c,$$ where $c$ is a real constant and $x,y$ are variables.
I've worked on the proof of this fact for a while, but could not get a clue. Could anyone give me a hint? Currently, I am trying to prove $|inf x-inf y|leq inf|x-y|$ if this is a correct direction.
real-analysis supremum-and-infimum
asked Jan 2 at 20:55
James WangJames Wang
50228
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add a comment |
$begingroup$
In the second last inequality of the proof of Theorem 6.17 (Rudin's baby analysis), I think he uses the fact that $$|x-y|leq c text{ implies } |inf x-inf y|leq c,$$ where $c$ is a real constant and $x,y$ are variables.
I've worked on the proof of this fact for a while, but could not get a clue. Could anyone give me a hint? Currently, I am trying to prove $|inf x-inf y|leq inf|x-y|$ if this is a correct direction.
real-analysis supremum-and-infimum
asked Jan 2 at 20:55
James WangJames Wang
50228
$endgroup$
$begingroup$
Can you tell us the set(s) to which $x,y$ belong to?
$endgroup$
– mathcounterexamples.net
Jan 2 at 21:06
$begingroup$
inf x is a dud. Say something intelligent like "inf range x" and state the range of x. It'd also help were you to state the orginal prkblem.
$endgroup$
– William Elliot
Jan 2 at 21:54
add a comment |
$begingroup$
In the second last inequality of the proof of Theorem 6.17 (Rudin's baby analysis), I think he uses the fact that $$|x-y|leq c text{ implies } |inf x-inf y|leq c,$$ where $c$ is a real constant and $x,y$ are variables.
I've worked on the proof of this fact for a while, but could not get a clue. Could anyone give me a hint? Currently, I am trying to prove $|inf x-inf y|leq inf|x-y|$ if this is a correct direction.
real-analysis supremum-and-infimum
asked Jan 2 at 20:55
James WangJames Wang
50228
$endgroup$
In the second last inequality of the proof of Theorem 6.17 (Rudin's baby analysis), I think he uses the fact that $$|x-y|leq c text{ implies } |inf x-inf y|leq c,$$ where $c$ is a real constant and $x,y$ are variables.
I've worked on the proof of this fact for a while, but could not get a clue. Could anyone give me a hint? Currently, I am trying to prove $|inf x-inf y|leq inf|x-y|$ if this is a correct direction.
real-analysis supremum-and-infimum
real-analysis supremum-and-infimum
asked Jan 2 at 20:55
James WangJames Wang
50228
asked Jan 2 at 20:55
James WangJames Wang
50228
asked Jan 2 at 20:55
James WangJames Wang
50228
asked Jan 2 at 20:55
James WangJames Wang
50228
asked Jan 2 at 20:55
James WangJames Wang
50228
50228
$begingroup$
Can you tell us the set(s) to which $x,y$ belong to?
$endgroup$
– mathcounterexamples.net
Jan 2 at 21:06
$begingroup$
inf x is a dud. Say something intelligent like "inf range x" and state the range of x. It'd also help were you to state the orginal prkblem.
$endgroup$
– William Elliot
Jan 2 at 21:54
add a comment |
$begingroup$
Can you tell us the set(s) to which $x,y$ belong to?
$endgroup$
– mathcounterexamples.net
Jan 2 at 21:06
$begingroup$
inf x is a dud. Say something intelligent like "inf range x" and state the range of x. It'd also help were you to state the orginal prkblem.
$endgroup$
– William Elliot
Jan 2 at 21:54
$begingroup$
Can you tell us the set(s) to which $x,y$ belong to?
$endgroup$
– mathcounterexamples.net
Jan 2 at 21:06
$begingroup$
Can you tell us the set(s) to which $x,y$ belong to?
$endgroup$
– mathcounterexamples.net
Jan 2 at 21:06
$begingroup$
inf x is a dud. Say something intelligent like "inf range x" and state the range of x. It'd also help were you to state the orginal prkblem.
$endgroup$
– William Elliot
Jan 2 at 21:54
$begingroup$
inf x is a dud. Say something intelligent like "inf range x" and state the range of x. It'd also help were you to state the orginal prkblem.
$endgroup$
– William Elliot
Jan 2 at 21:54
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I assume you mean that $x$ and $y$ belong to subsets $X$ and $Y$, respectively, of $mathbb R$.
Let $epsilon>0$. By definition of $inf$, there are $x_0in X; y_0in Y$ such that
$inf X>x_0-epsilon$ and $inf Y>y_0-epsilon.$
And an application of the triangle inequality gives
$|inf X-inf Y|le |inf X-x_0|+|x_0-y_0|+|y_0-inf X|le 2epsilon+c.$
As $epsilon$ is arbitrary, the result follows.
answered Jan 3 at 0:09
MatematletaMatematleta
12k21020
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add a comment |
$begingroup$
It feels like it makes more sense to treat $x$ and $y$ as (real-valued) functions of some parameter $t$; there isn't anything interesting here unless $x$ and $y$ are related to each other somehow.
WLOG, $inf yle inf x$. Now, what can we do with those infimums? Approach one of them - we can find some $t$ such that $x$ is within $epsilon$ of its infimum, or maybe some $t$ such that $y$ is within $epsilon$ of its infimum. Which one? Try both, and see which one leads somewhere.
answered Jan 2 at 21:36
jmerryjmerry
16.9k11633
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add a comment |
$begingroup$
I think what you are trying to prove is the following: if $A,B subset mathbb R$ and $|a-b| leq c$ whenever $ain A$ and $b in B$ then $|inf , A-inf, B| leq c$. To prove this note that $aleq b+c$ whenever $ain A$ and $b in B$. This implies $inf, A leq b+c$ or $inf, A -cleq b$ whenever $b in B$. In turn, this implies $inf, A -cleq inf, B$ or $inf, Aleq inf, B+c$. Similarly, $inf, Bleq inf, A+c$. Combining these two we get $|inf , A-inf, B| leq c$.
answered Jan 3 at 0:01
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I assume you mean that $x$ and $y$ belong to subsets $X$ and $Y$, respectively, of $mathbb R$.
Let $epsilon>0$. By definition of $inf$, there are $x_0in X; y_0in Y$ such that
$inf X>x_0-epsilon$ and $inf Y>y_0-epsilon.$
And an application of the triangle inequality gives
$|inf X-inf Y|le |inf X-x_0|+|x_0-y_0|+|y_0-inf X|le 2epsilon+c.$
As $epsilon$ is arbitrary, the result follows.
answered Jan 3 at 0:09
MatematletaMatematleta
12k21020
$endgroup$
add a comment |
$begingroup$
I assume you mean that $x$ and $y$ belong to subsets $X$ and $Y$, respectively, of $mathbb R$.
Let $epsilon>0$. By definition of $inf$, there are $x_0in X; y_0in Y$ such that
$inf X>x_0-epsilon$ and $inf Y>y_0-epsilon.$
And an application of the triangle inequality gives
$|inf X-inf Y|le |inf X-x_0|+|x_0-y_0|+|y_0-inf X|le 2epsilon+c.$
As $epsilon$ is arbitrary, the result follows.
answered Jan 3 at 0:09
MatematletaMatematleta
12k21020
$endgroup$
add a comment |
$begingroup$
I assume you mean that $x$ and $y$ belong to subsets $X$ and $Y$, respectively, of $mathbb R$.
Let $epsilon>0$. By definition of $inf$, there are $x_0in X; y_0in Y$ such that
$inf X>x_0-epsilon$ and $inf Y>y_0-epsilon.$
And an application of the triangle inequality gives
$|inf X-inf Y|le |inf X-x_0|+|x_0-y_0|+|y_0-inf X|le 2epsilon+c.$
As $epsilon$ is arbitrary, the result follows.
answered Jan 3 at 0:09
MatematletaMatematleta
12k21020
$endgroup$
I assume you mean that $x$ and $y$ belong to subsets $X$ and $Y$, respectively, of $mathbb R$.
Let $epsilon>0$. By definition of $inf$, there are $x_0in X; y_0in Y$ such that
$inf X>x_0-epsilon$ and $inf Y>y_0-epsilon.$
And an application of the triangle inequality gives
$|inf X-inf Y|le |inf X-x_0|+|x_0-y_0|+|y_0-inf X|le 2epsilon+c.$
As $epsilon$ is arbitrary, the result follows.
answered Jan 3 at 0:09
MatematletaMatematleta
12k21020
answered Jan 3 at 0:09
MatematletaMatematleta
12k21020
answered Jan 3 at 0:09
MatematletaMatematleta
12k21020
answered Jan 3 at 0:09
MatematletaMatematleta
12k21020
12k21020
add a comment |
add a comment |
$begingroup$
It feels like it makes more sense to treat $x$ and $y$ as (real-valued) functions of some parameter $t$; there isn't anything interesting here unless $x$ and $y$ are related to each other somehow.
WLOG, $inf yle inf x$. Now, what can we do with those infimums? Approach one of them - we can find some $t$ such that $x$ is within $epsilon$ of its infimum, or maybe some $t$ such that $y$ is within $epsilon$ of its infimum. Which one? Try both, and see which one leads somewhere.
answered Jan 2 at 21:36
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
It feels like it makes more sense to treat $x$ and $y$ as (real-valued) functions of some parameter $t$; there isn't anything interesting here unless $x$ and $y$ are related to each other somehow.
WLOG, $inf yle inf x$. Now, what can we do with those infimums? Approach one of them - we can find some $t$ such that $x$ is within $epsilon$ of its infimum, or maybe some $t$ such that $y$ is within $epsilon$ of its infimum. Which one? Try both, and see which one leads somewhere.
answered Jan 2 at 21:36
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
It feels like it makes more sense to treat $x$ and $y$ as (real-valued) functions of some parameter $t$; there isn't anything interesting here unless $x$ and $y$ are related to each other somehow.
WLOG, $inf yle inf x$. Now, what can we do with those infimums? Approach one of them - we can find some $t$ such that $x$ is within $epsilon$ of its infimum, or maybe some $t$ such that $y$ is within $epsilon$ of its infimum. Which one? Try both, and see which one leads somewhere.
answered Jan 2 at 21:36
jmerryjmerry
16.9k11633
$endgroup$
It feels like it makes more sense to treat $x$ and $y$ as (real-valued) functions of some parameter $t$; there isn't anything interesting here unless $x$ and $y$ are related to each other somehow.
WLOG, $inf yle inf x$. Now, what can we do with those infimums? Approach one of them - we can find some $t$ such that $x$ is within $epsilon$ of its infimum, or maybe some $t$ such that $y$ is within $epsilon$ of its infimum. Which one? Try both, and see which one leads somewhere.
answered Jan 2 at 21:36
jmerryjmerry
16.9k11633
answered Jan 2 at 21:36
jmerryjmerry
16.9k11633
answered Jan 2 at 21:36
jmerryjmerry
16.9k11633
answered Jan 2 at 21:36
jmerryjmerry
16.9k11633
16.9k11633
add a comment |
add a comment |
$begingroup$
I think what you are trying to prove is the following: if $A,B subset mathbb R$ and $|a-b| leq c$ whenever $ain A$ and $b in B$ then $|inf , A-inf, B| leq c$. To prove this note that $aleq b+c$ whenever $ain A$ and $b in B$. This implies $inf, A leq b+c$ or $inf, A -cleq b$ whenever $b in B$. In turn, this implies $inf, A -cleq inf, B$ or $inf, Aleq inf, B+c$. Similarly, $inf, Bleq inf, A+c$. Combining these two we get $|inf , A-inf, B| leq c$.
answered Jan 3 at 0:01
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
$endgroup$
add a comment |
$begingroup$
I think what you are trying to prove is the following: if $A,B subset mathbb R$ and $|a-b| leq c$ whenever $ain A$ and $b in B$ then $|inf , A-inf, B| leq c$. To prove this note that $aleq b+c$ whenever $ain A$ and $b in B$. This implies $inf, A leq b+c$ or $inf, A -cleq b$ whenever $b in B$. In turn, this implies $inf, A -cleq inf, B$ or $inf, Aleq inf, B+c$. Similarly, $inf, Bleq inf, A+c$. Combining these two we get $|inf , A-inf, B| leq c$.
answered Jan 3 at 0:01
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
$endgroup$
add a comment |
$begingroup$
I think what you are trying to prove is the following: if $A,B subset mathbb R$ and $|a-b| leq c$ whenever $ain A$ and $b in B$ then $|inf , A-inf, B| leq c$. To prove this note that $aleq b+c$ whenever $ain A$ and $b in B$. This implies $inf, A leq b+c$ or $inf, A -cleq b$ whenever $b in B$. In turn, this implies $inf, A -cleq inf, B$ or $inf, Aleq inf, B+c$. Similarly, $inf, Bleq inf, A+c$. Combining these two we get $|inf , A-inf, B| leq c$.
answered Jan 3 at 0:01
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
$endgroup$
I think what you are trying to prove is the following: if $A,B subset mathbb R$ and $|a-b| leq c$ whenever $ain A$ and $b in B$ then $|inf , A-inf, B| leq c$. To prove this note that $aleq b+c$ whenever $ain A$ and $b in B$. This implies $inf, A leq b+c$ or $inf, A -cleq b$ whenever $b in B$. In turn, this implies $inf, A -cleq inf, B$ or $inf, Aleq inf, B+c$. Similarly, $inf, Bleq inf, A+c$. Combining these two we get $|inf , A-inf, B| leq c$.
answered Jan 3 at 0:01
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
answered Jan 3 at 0:01
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
answered Jan 3 at 0:01
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
answered Jan 3 at 0:01
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
71.4k53170
add a comment |
add a comment |
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$begingroup$
Can you tell us the set(s) to which $x,y$ belong to?
$endgroup$
– mathcounterexamples.net
Jan 2 at 21:06
$begingroup$
inf x is a dud. Say something intelligent like "inf range x" and state the range of x. It'd also help were you to state the orginal prkblem.
$endgroup$
– William Elliot
Jan 2 at 21:54