Mixing
Piecewise Functions and Supremums
$begingroup$
Let $f(x) = begin{cases}
x, &x < a\
x + 3, & x geq a
end{cases}$ $hspace{1cm}$ and $A = left { f(x): x < aright }$
Is $text{sup} A= f(a)$?
I suspect not because it is NOT the case that for any arbitrary positive number $f(a) - epsilon < f(x)$ for $x in A$. In other words, there exists epsilon such that $f(a) - epsilon geq f(x)$, thus contradicting that $text{sup} A= f(a)$.
But then what is sup $A$?
real-analysis
asked Jan 30 at 17:46
user_hello1user_hello1
987
$endgroup$
add a comment |
$begingroup$
Let $f(x) = begin{cases}
x, &x < a\
x + 3, & x geq a
end{cases}$ $hspace{1cm}$ and $A = left { f(x): x < aright }$
Is $text{sup} A= f(a)$?
I suspect not because it is NOT the case that for any arbitrary positive number $f(a) - epsilon < f(x)$ for $x in A$. In other words, there exists epsilon such that $f(a) - epsilon geq f(x)$, thus contradicting that $text{sup} A= f(a)$.
But then what is sup $A$?
real-analysis
asked Jan 30 at 17:46
user_hello1user_hello1
987
$endgroup$
1
$begingroup$
the supremum would just be $a$ itself. It's an upper bound, and the smallest one, but it just leaves a "hole" in the graph
$endgroup$
– rubikscube09
Jan 30 at 17:49
add a comment |
$begingroup$
Let $f(x) = begin{cases}
x, &x < a\
x + 3, & x geq a
end{cases}$ $hspace{1cm}$ and $A = left { f(x): x < aright }$
Is $text{sup} A= f(a)$?
I suspect not because it is NOT the case that for any arbitrary positive number $f(a) - epsilon < f(x)$ for $x in A$. In other words, there exists epsilon such that $f(a) - epsilon geq f(x)$, thus contradicting that $text{sup} A= f(a)$.
But then what is sup $A$?
real-analysis
asked Jan 30 at 17:46
user_hello1user_hello1
987
$endgroup$
Let $f(x) = begin{cases}
x, &x < a\
x + 3, & x geq a
end{cases}$ $hspace{1cm}$ and $A = left { f(x): x < aright }$
Is $text{sup} A= f(a)$?
I suspect not because it is NOT the case that for any arbitrary positive number $f(a) - epsilon < f(x)$ for $x in A$. In other words, there exists epsilon such that $f(a) - epsilon geq f(x)$, thus contradicting that $text{sup} A= f(a)$.
But then what is sup $A$?
real-analysis
real-analysis
asked Jan 30 at 17:46
user_hello1user_hello1
987
asked Jan 30 at 17:46
user_hello1user_hello1
987
asked Jan 30 at 17:46
user_hello1user_hello1
987
asked Jan 30 at 17:46
user_hello1user_hello1
987
asked Jan 30 at 17:46
user_hello1user_hello1
987
987
1
$begingroup$
the supremum would just be $a$ itself. It's an upper bound, and the smallest one, but it just leaves a "hole" in the graph
$endgroup$
– rubikscube09
Jan 30 at 17:49
add a comment |
1
$begingroup$
the supremum would just be $a$ itself. It's an upper bound, and the smallest one, but it just leaves a "hole" in the graph
$endgroup$
– rubikscube09
Jan 30 at 17:49
1
1
$begingroup$
the supremum would just be $a$ itself. It's an upper bound, and the smallest one, but it just leaves a "hole" in the graph
$endgroup$
– rubikscube09
Jan 30 at 17:49
$begingroup$
the supremum would just be $a$ itself. It's an upper bound, and the smallest one, but it just leaves a "hole" in the graph
$endgroup$
– rubikscube09
Jan 30 at 17:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $x < a$ , $f(x) = x$.
So basically $A = {f(x) : x <a } = { x : x <a }$, and the supremum of this set is a.
answered Jan 30 at 20:35
Daphna KeidarDaphna Keidar
34417
$endgroup$
add a comment |
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$begingroup$
For $x < a$ , $f(x) = x$.
So basically $A = {f(x) : x <a } = { x : x <a }$, and the supremum of this set is a.
answered Jan 30 at 20:35
Daphna KeidarDaphna Keidar
34417
$endgroup$
add a comment |
$begingroup$
For $x < a$ , $f(x) = x$.
So basically $A = {f(x) : x <a } = { x : x <a }$, and the supremum of this set is a.
answered Jan 30 at 20:35
Daphna KeidarDaphna Keidar
34417
$endgroup$
add a comment |
$begingroup$
For $x < a$ , $f(x) = x$.
So basically $A = {f(x) : x <a } = { x : x <a }$, and the supremum of this set is a.
answered Jan 30 at 20:35
Daphna KeidarDaphna Keidar
34417
$endgroup$
For $x < a$ , $f(x) = x$.
So basically $A = {f(x) : x <a } = { x : x <a }$, and the supremum of this set is a.
answered Jan 30 at 20:35
Daphna KeidarDaphna Keidar
34417
answered Jan 30 at 20:35
Daphna KeidarDaphna Keidar
34417
answered Jan 30 at 20:35
Daphna KeidarDaphna Keidar
34417
answered Jan 30 at 20:35
Daphna KeidarDaphna Keidar
34417
34417
add a comment |
add a comment |
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$begingroup$
the supremum would just be $a$ itself. It's an upper bound, and the smallest one, but it just leaves a "hole" in the graph
$endgroup$
– rubikscube09
Jan 30 at 17:49