Baby Rudin's exercise 2.21
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Here's the exercise:
Let $A$ and $B$ be separated subsets of some $mathbbR^k$, suppose $a in A$, $b in B$, and define:
$$p(t) = (1-t)a + tb$$
for $t in mathbbR^1$. Put $A_0 = p^-1 (A)$, $B_0 = p^-1 (B)$ [Thus $t in A_0 iff p(t) in A$]
My problem is that I don't get just what is $p(t)$. Is it a mapping? If so, then what is $p(A)$? Is it a set?
I'm sorry if it sounds stupid or too elementary but I simply don't understand what is this $p(t)$ Rudin just defined. It resembles a property of convexity - but it accepts any real $t$, not just a real $t$ such that $0<t<1$.
I think this has something to do with convexity in finite dimensions, but I'm not sure.
real-analysis convex-analysis
asked Aug 26 at 19:31
Pedro Cavalcante Oliveira
205
 |Â
show 2 more comments
up vote
0
down vote
favorite
Here's the exercise:
Let $A$ and $B$ be separated subsets of some $mathbbR^k$, suppose $a in A$, $b in B$, and define:
$$p(t) = (1-t)a + tb$$
for $t in mathbbR^1$. Put $A_0 = p^-1 (A)$, $B_0 = p^-1 (B)$ [Thus $t in A_0 iff p(t) in A$]
My problem is that I don't get just what is $p(t)$. Is it a mapping? If so, then what is $p(A)$? Is it a set?
I'm sorry if it sounds stupid or too elementary but I simply don't understand what is this $p(t)$ Rudin just defined. It resembles a property of convexity - but it accepts any real $t$, not just a real $t$ such that $0<t<1$.
I think this has something to do with convexity in finite dimensions, but I'm not sure.
real-analysis convex-analysis
asked Aug 26 at 19:31
Pedro Cavalcante Oliveira
205
1
Yes, $tmapsto p(t)$ is a mapping. There is no $p(A)$, only $p^-1(A)$.
– Dietrich Burde
Aug 26 at 19:34
$p$ is a function that maps $t in mathbb R$ into $p(t) in mathbb R^k$. $p^-1(A)$ is the inverse image of $A$ under $p$. As @Dietrich points out below, $p(A)$ is meaningless, because $A subset mathbb R^k$ but $p$ is defined only on $mathbb R$.
– Theoretical Economist
Aug 26 at 19:34
@DietrichBurde you're right, I wasn't reading the question properly. Will edit my comment.
– Theoretical Economist
Aug 26 at 19:37
I researched for this question and couldn't find it, my bad. Thanks for the link, Dietrich.
– Pedro Cavalcante Oliveira
Aug 26 at 19:37
1
Have a look at file: minds.wisconsin.edu/handle/1793/67009
– Bruce
Aug 26 at 19:41
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Here's the exercise:
Let $A$ and $B$ be separated subsets of some $mathbbR^k$, suppose $a in A$, $b in B$, and define:
$$p(t) = (1-t)a + tb$$
for $t in mathbbR^1$. Put $A_0 = p^-1 (A)$, $B_0 = p^-1 (B)$ [Thus $t in A_0 iff p(t) in A$]
My problem is that I don't get just what is $p(t)$. Is it a mapping? If so, then what is $p(A)$? Is it a set?
I'm sorry if it sounds stupid or too elementary but I simply don't understand what is this $p(t)$ Rudin just defined. It resembles a property of convexity - but it accepts any real $t$, not just a real $t$ such that $0<t<1$.
I think this has something to do with convexity in finite dimensions, but I'm not sure.
real-analysis convex-analysis
asked Aug 26 at 19:31
Pedro Cavalcante Oliveira
205
Here's the exercise:
Let $A$ and $B$ be separated subsets of some $mathbbR^k$, suppose $a in A$, $b in B$, and define:
$$p(t) = (1-t)a + tb$$
for $t in mathbbR^1$. Put $A_0 = p^-1 (A)$, $B_0 = p^-1 (B)$ [Thus $t in A_0 iff p(t) in A$]
My problem is that I don't get just what is $p(t)$. Is it a mapping? If so, then what is $p(A)$? Is it a set?
I'm sorry if it sounds stupid or too elementary but I simply don't understand what is this $p(t)$ Rudin just defined. It resembles a property of convexity - but it accepts any real $t$, not just a real $t$ such that $0<t<1$.
I think this has something to do with convexity in finite dimensions, but I'm not sure.
real-analysis convex-analysis
asked Aug 26 at 19:31
Pedro Cavalcante Oliveira
205
edited Aug 26 at 20:25
asked Aug 26 at 19:31
Pedro Cavalcante Oliveira
205
asked Aug 26 at 19:31
Pedro Cavalcante Oliveira
205
asked Aug 26 at 19:31
Pedro Cavalcante Oliveira
205
205
1
Yes, $tmapsto p(t)$ is a mapping. There is no $p(A)$, only $p^-1(A)$.
– Dietrich Burde
Aug 26 at 19:34
$p$ is a function that maps $t in mathbb R$ into $p(t) in mathbb R^k$. $p^-1(A)$ is the inverse image of $A$ under $p$. As @Dietrich points out below, $p(A)$ is meaningless, because $A subset mathbb R^k$ but $p$ is defined only on $mathbb R$.
– Theoretical Economist
Aug 26 at 19:34
@DietrichBurde you're right, I wasn't reading the question properly. Will edit my comment.
– Theoretical Economist
Aug 26 at 19:37
I researched for this question and couldn't find it, my bad. Thanks for the link, Dietrich.
– Pedro Cavalcante Oliveira
Aug 26 at 19:37
1
Have a look at file: minds.wisconsin.edu/handle/1793/67009
– Bruce
Aug 26 at 19:41
 |Â
show 2 more comments
1
Yes, $tmapsto p(t)$ is a mapping. There is no $p(A)$, only $p^-1(A)$.
– Dietrich Burde
Aug 26 at 19:34
$p$ is a function that maps $t in mathbb R$ into $p(t) in mathbb R^k$. $p^-1(A)$ is the inverse image of $A$ under $p$. As @Dietrich points out below, $p(A)$ is meaningless, because $A subset mathbb R^k$ but $p$ is defined only on $mathbb R$.
– Theoretical Economist
Aug 26 at 19:34
@DietrichBurde you're right, I wasn't reading the question properly. Will edit my comment.
– Theoretical Economist
Aug 26 at 19:37
I researched for this question and couldn't find it, my bad. Thanks for the link, Dietrich.
– Pedro Cavalcante Oliveira
Aug 26 at 19:37
1
Have a look at file: minds.wisconsin.edu/handle/1793/67009
– Bruce
Aug 26 at 19:41
1
1
Yes, $tmapsto p(t)$ is a mapping. There is no $p(A)$, only $p^-1(A)$.
– Dietrich Burde
Aug 26 at 19:34
Yes, $tmapsto p(t)$ is a mapping. There is no $p(A)$, only $p^-1(A)$.
– Dietrich Burde
Aug 26 at 19:34
$p$ is a function that maps $t in mathbb R$ into $p(t) in mathbb R^k$. $p^-1(A)$ is the inverse image of $A$ under $p$. As @Dietrich points out below, $p(A)$ is meaningless, because $A subset mathbb R^k$ but $p$ is defined only on $mathbb R$.
– Theoretical Economist
Aug 26 at 19:34
$p$ is a function that maps $t in mathbb R$ into $p(t) in mathbb R^k$. $p^-1(A)$ is the inverse image of $A$ under $p$. As @Dietrich points out below, $p(A)$ is meaningless, because $A subset mathbb R^k$ but $p$ is defined only on $mathbb R$.
– Theoretical Economist
Aug 26 at 19:34
@DietrichBurde you're right, I wasn't reading the question properly. Will edit my comment.
– Theoretical Economist
Aug 26 at 19:37
@DietrichBurde you're right, I wasn't reading the question properly. Will edit my comment.
– Theoretical Economist
Aug 26 at 19:37
I researched for this question and couldn't find it, my bad. Thanks for the link, Dietrich.
– Pedro Cavalcante Oliveira
Aug 26 at 19:37
I researched for this question and couldn't find it, my bad. Thanks for the link, Dietrich.
– Pedro Cavalcante Oliveira
Aug 26 at 19:37
1
1
Have a look at file: minds.wisconsin.edu/handle/1793/67009
– Bruce
Aug 26 at 19:41
Have a look at file: minds.wisconsin.edu/handle/1793/67009
– Bruce
Aug 26 at 19:41
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
$p: mathbbR to mathbbR^n: t mapsto p(t)$ is a map.
Geometrically, the graph of the function is the line between the points a and b.
Btw, I dislike Rudin's approach to show that convex sets are connected. It's easier and more general to show that pathconnected sets are connected and then it's trivial to see that convex sets are connected (using the map that Rudin provides).
answered Aug 26 at 21:05
Math_QED
6,53931345
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$p: mathbbR to mathbbR^n: t mapsto p(t)$ is a map.
Geometrically, the graph of the function is the line between the points a and b.
Btw, I dislike Rudin's approach to show that convex sets are connected. It's easier and more general to show that pathconnected sets are connected and then it's trivial to see that convex sets are connected (using the map that Rudin provides).
answered Aug 26 at 21:05
Math_QED
6,53931345
add a comment |Â
up vote
1
down vote
accepted
$p: mathbbR to mathbbR^n: t mapsto p(t)$ is a map.
Geometrically, the graph of the function is the line between the points a and b.
Btw, I dislike Rudin's approach to show that convex sets are connected. It's easier and more general to show that pathconnected sets are connected and then it's trivial to see that convex sets are connected (using the map that Rudin provides).
answered Aug 26 at 21:05
Math_QED
6,53931345
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$p: mathbbR to mathbbR^n: t mapsto p(t)$ is a map.
Geometrically, the graph of the function is the line between the points a and b.
Btw, I dislike Rudin's approach to show that convex sets are connected. It's easier and more general to show that pathconnected sets are connected and then it's trivial to see that convex sets are connected (using the map that Rudin provides).
answered Aug 26 at 21:05
Math_QED
6,53931345
$p: mathbbR to mathbbR^n: t mapsto p(t)$ is a map.
Geometrically, the graph of the function is the line between the points a and b.
Btw, I dislike Rudin's approach to show that convex sets are connected. It's easier and more general to show that pathconnected sets are connected and then it's trivial to see that convex sets are connected (using the map that Rudin provides).
answered Aug 26 at 21:05
Math_QED
6,53931345
edited Aug 26 at 21:14
answered Aug 26 at 21:05
Math_QED
6,53931345
answered Aug 26 at 21:05
Math_QED
6,53931345
answered Aug 26 at 21:05
Math_QED
6,53931345
6,53931345
add a comment |Â
add a comment |Â
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1
Yes, $tmapsto p(t)$ is a mapping. There is no $p(A)$, only $p^-1(A)$.
– Dietrich Burde
Aug 26 at 19:34
$p$ is a function that maps $t in mathbb R$ into $p(t) in mathbb R^k$. $p^-1(A)$ is the inverse image of $A$ under $p$. As @Dietrich points out below, $p(A)$ is meaningless, because $A subset mathbb R^k$ but $p$ is defined only on $mathbb R$.
– Theoretical Economist
Aug 26 at 19:34
@DietrichBurde you're right, I wasn't reading the question properly. Will edit my comment.
– Theoretical Economist
Aug 26 at 19:37
I researched for this question and couldn't find it, my bad. Thanks for the link, Dietrich.
– Pedro Cavalcante Oliveira
Aug 26 at 19:37
1
Have a look at file: minds.wisconsin.edu/handle/1793/67009
– Bruce
Aug 26 at 19:41