Mixing
Change of quantifiers in the definition of continuity.
$begingroup$
Let $f(x)$ be a continuous function on $Bbb{R}$ and $a in Bbb{R}$
$forall epsilon>0$ we define the set $$S_{epsilon}={delta>0|forall x:0<|x-a|<delta Rightarrow 0<|f(x)-f(a)|<epsilon}$$
Assume that $forall epsilon>0$ we have $S_{epsilon}=(0,+infty)$
What can someone conclude about $f(x)$?
I believe that $f$ is constant $f(a)$
Assume that $exists x_0 in Bbb{R}$ such that $|f(x_0)-f(a)|>0$
Put $epsilon_0=|f(x_0)-f(a)|>0$
Then we have that $S_{epsilon_0}=(0,+infty)$
We choose $delta_0>0$ big enough such that $x_0 in (a-delta_0,a+delta_0)$
Then we can easily see that $delta_0 notin S_{epsilon_0}Rightarrow S_{epsilon_0} neq (0,+infty)$ which is a contradiction.
Thus $f(x)=f(a), forall x in Bbb{R}$.
Is my proof correct?
Thank you in advance.
real-analysis proof-verification continuity epsilon-delta
asked Jan 16 at 12:09
Marios GretsasMarios Gretsas
8,48011437
$endgroup$
add a comment |
$begingroup$
Let $f(x)$ be a continuous function on $Bbb{R}$ and $a in Bbb{R}$
$forall epsilon>0$ we define the set $$S_{epsilon}={delta>0|forall x:0<|x-a|<delta Rightarrow 0<|f(x)-f(a)|<epsilon}$$
Assume that $forall epsilon>0$ we have $S_{epsilon}=(0,+infty)$
What can someone conclude about $f(x)$?
I believe that $f$ is constant $f(a)$
Assume that $exists x_0 in Bbb{R}$ such that $|f(x_0)-f(a)|>0$
Put $epsilon_0=|f(x_0)-f(a)|>0$
Then we have that $S_{epsilon_0}=(0,+infty)$
We choose $delta_0>0$ big enough such that $x_0 in (a-delta_0,a+delta_0)$
Then we can easily see that $delta_0 notin S_{epsilon_0}Rightarrow S_{epsilon_0} neq (0,+infty)$ which is a contradiction.
Thus $f(x)=f(a), forall x in Bbb{R}$.
Is my proof correct?
Thank you in advance.
real-analysis proof-verification continuity epsilon-delta
asked Jan 16 at 12:09
Marios GretsasMarios Gretsas
8,48011437
$endgroup$
2
$begingroup$
Yes, it is correct.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 12:16
$begingroup$
Nice..Thank you.:)
$endgroup$
– Marios Gretsas
Jan 16 at 12:17
$begingroup$
In the definition of S are a and x supposed to be constants?
$endgroup$
– William Elliot
Jan 16 at 12:20
$begingroup$
Only $a$ is a constant..$x$ is not from hypothesis that $S_{epsilon}=(0,+infty)$
$endgroup$
– Marios Gretsas
Jan 16 at 12:22
$begingroup$
@MariosGretsas. Then the definition of S is missing a quantifier.
$endgroup$
– William Elliot
Jan 16 at 21:37
add a comment |
$begingroup$
Let $f(x)$ be a continuous function on $Bbb{R}$ and $a in Bbb{R}$
$forall epsilon>0$ we define the set $$S_{epsilon}={delta>0|forall x:0<|x-a|<delta Rightarrow 0<|f(x)-f(a)|<epsilon}$$
Assume that $forall epsilon>0$ we have $S_{epsilon}=(0,+infty)$
What can someone conclude about $f(x)$?
I believe that $f$ is constant $f(a)$
Assume that $exists x_0 in Bbb{R}$ such that $|f(x_0)-f(a)|>0$
Put $epsilon_0=|f(x_0)-f(a)|>0$
Then we have that $S_{epsilon_0}=(0,+infty)$
We choose $delta_0>0$ big enough such that $x_0 in (a-delta_0,a+delta_0)$
Then we can easily see that $delta_0 notin S_{epsilon_0}Rightarrow S_{epsilon_0} neq (0,+infty)$ which is a contradiction.
Thus $f(x)=f(a), forall x in Bbb{R}$.
Is my proof correct?
Thank you in advance.
real-analysis proof-verification continuity epsilon-delta
asked Jan 16 at 12:09
Marios GretsasMarios Gretsas
8,48011437
$endgroup$
Let $f(x)$ be a continuous function on $Bbb{R}$ and $a in Bbb{R}$
$forall epsilon>0$ we define the set $$S_{epsilon}={delta>0|forall x:0<|x-a|<delta Rightarrow 0<|f(x)-f(a)|<epsilon}$$
Assume that $forall epsilon>0$ we have $S_{epsilon}=(0,+infty)$
What can someone conclude about $f(x)$?
I believe that $f$ is constant $f(a)$
Assume that $exists x_0 in Bbb{R}$ such that $|f(x_0)-f(a)|>0$
Put $epsilon_0=|f(x_0)-f(a)|>0$
Then we have that $S_{epsilon_0}=(0,+infty)$
We choose $delta_0>0$ big enough such that $x_0 in (a-delta_0,a+delta_0)$
Then we can easily see that $delta_0 notin S_{epsilon_0}Rightarrow S_{epsilon_0} neq (0,+infty)$ which is a contradiction.
Thus $f(x)=f(a), forall x in Bbb{R}$.
Is my proof correct?
Thank you in advance.
real-analysis proof-verification continuity epsilon-delta
real-analysis proof-verification continuity epsilon-delta
asked Jan 16 at 12:09
Marios GretsasMarios Gretsas
8,48011437
asked Jan 16 at 12:09
Marios GretsasMarios Gretsas
8,48011437
edited Jan 17 at 9:31
Marios Gretsas
asked Jan 16 at 12:09
Marios GretsasMarios Gretsas
8,48011437
asked Jan 16 at 12:09
Marios GretsasMarios Gretsas
8,48011437
asked Jan 16 at 12:09
Marios GretsasMarios Gretsas
8,48011437
8,48011437
2
$begingroup$
Yes, it is correct.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 12:16
$begingroup$
Nice..Thank you.:)
$endgroup$
– Marios Gretsas
Jan 16 at 12:17
$begingroup$
In the definition of S are a and x supposed to be constants?
$endgroup$
– William Elliot
Jan 16 at 12:20
$begingroup$
Only $a$ is a constant..$x$ is not from hypothesis that $S_{epsilon}=(0,+infty)$
$endgroup$
– Marios Gretsas
Jan 16 at 12:22
$begingroup$
@MariosGretsas. Then the definition of S is missing a quantifier.
$endgroup$
– William Elliot
Jan 16 at 21:37
add a comment |
2
$begingroup$
Yes, it is correct.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 12:16
$begingroup$
Nice..Thank you.:)
$endgroup$
– Marios Gretsas
Jan 16 at 12:17
$begingroup$
In the definition of S are a and x supposed to be constants?
$endgroup$
– William Elliot
Jan 16 at 12:20
$begingroup$
Only $a$ is a constant..$x$ is not from hypothesis that $S_{epsilon}=(0,+infty)$
$endgroup$
– Marios Gretsas
Jan 16 at 12:22
$begingroup$
@MariosGretsas. Then the definition of S is missing a quantifier.
$endgroup$
– William Elliot
Jan 16 at 21:37
2
2
$begingroup$
Yes, it is correct.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 12:16
$begingroup$
Yes, it is correct.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 12:16
$begingroup$
Nice..Thank you.:)
$endgroup$
– Marios Gretsas
Jan 16 at 12:17
$begingroup$
Nice..Thank you.:)
$endgroup$
– Marios Gretsas
Jan 16 at 12:17
$begingroup$
In the definition of S are a and x supposed to be constants?
$endgroup$
– William Elliot
Jan 16 at 12:20
$begingroup$
In the definition of S are a and x supposed to be constants?
$endgroup$
– William Elliot
Jan 16 at 12:20
$begingroup$
Only $a$ is a constant..$x$ is not from hypothesis that $S_{epsilon}=(0,+infty)$
$endgroup$
– Marios Gretsas
Jan 16 at 12:22
$begingroup$
Only $a$ is a constant..$x$ is not from hypothesis that $S_{epsilon}=(0,+infty)$
$endgroup$
– Marios Gretsas
Jan 16 at 12:22
$begingroup$
@MariosGretsas. Then the definition of S is missing a quantifier.
$endgroup$
– William Elliot
Jan 16 at 21:37
$begingroup$
@MariosGretsas. Then the definition of S is missing a quantifier.
$endgroup$
– William Elliot
Jan 16 at 21:37
add a comment |
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2
$begingroup$
Yes, it is correct.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 12:16
$begingroup$
Nice..Thank you.:)
$endgroup$
– Marios Gretsas
Jan 16 at 12:17
$begingroup$
In the definition of S are a and x supposed to be constants?
$endgroup$
– William Elliot
Jan 16 at 12:20
$begingroup$
Only $a$ is a constant..$x$ is not from hypothesis that $S_{epsilon}=(0,+infty)$
$endgroup$
– Marios Gretsas
Jan 16 at 12:22
$begingroup$
@MariosGretsas. Then the definition of S is missing a quantifier.
$endgroup$
– William Elliot
Jan 16 at 21:37