Show that the indicator function is discontinuous at every point $a in mathbb{R}$
$begingroup$
Let $S subset mathbb{R^n}$ and let $f$ be a function from $S$ into $mathbb{R^m}$. A function $f$ is continuous at $a$ if $forall epsilon > 0, exists delta > 0$ such that if $bigg| x - a bigg| < delta rightarrow bigg|f(x) - f(a) bigg| < epsilon$ For
Prove that the indicator function: $$f(x) = begin {cases} 0 & xnotin mathbb{Q}\ 1 & x in mathbb{Q} end{cases} $$
is discontinuous at every point $a in mathbb{R}$.
My professor assigned this exercise as practice for us to do. Specifically he asked us to show "formally" that it is the case that the indicator function is discontinuous. I understand exactly why it is discontinuous, I would like to know if the manner in which I proved it is "mathematically rigorous".
Proof:
By the Archimedian properties of $mathbb{R}$, for all $x,y in mathbb{Q}$, there exists $w notin mathbb{Q}$ such that $x < w < y$. As well for all $q_1, q_2 notin mathbb{Q}$, there exists $r in mathbb{Q}$ such that $q_1 < r < q_2$
With this being the case there are two cases to examine:
$x in mathbb{Q}, a notin mathbb{Q}$ AND
$a in mathbb{Q}, x notin mathbb{Q}$
Without loss of generality let's suppose:
$x in mathbb{Q}, a notin mathbb{Q}$
The other follows directly.
Let $epsilon = 0$, $delta > 0$, and let $x = frac{delta}{2} + a$
Therefore by definition: $$ bigg| x - a bigg| = bigg| (frac{delta}{2} + a) - a bigg| = frac{delta}{2} < delta$$
This means: $$bigg|f(x) - f(a) bigg| = |1 - 0| = 1 > epsilon = 0$$
I feel that this is sufficient because by invoking the Archimedian properties I take care of the whole issue of the denseness with $mathbb{Q}$. But this is why I am asking the question here. Could I get feedback?
real-analysis proof-verification
asked Dec 3 '18 at 22:19
dc3rddc3rd
1,47011137
$endgroup$
add a comment |
$begingroup$
Let $S subset mathbb{R^n}$ and let $f$ be a function from $S$ into $mathbb{R^m}$. A function $f$ is continuous at $a$ if $forall epsilon > 0, exists delta > 0$ such that if $bigg| x - a bigg| < delta rightarrow bigg|f(x) - f(a) bigg| < epsilon$ For
Prove that the indicator function: $$f(x) = begin {cases} 0 & xnotin mathbb{Q}\ 1 & x in mathbb{Q} end{cases} $$
is discontinuous at every point $a in mathbb{R}$.
My professor assigned this exercise as practice for us to do. Specifically he asked us to show "formally" that it is the case that the indicator function is discontinuous. I understand exactly why it is discontinuous, I would like to know if the manner in which I proved it is "mathematically rigorous".
Proof:
By the Archimedian properties of $mathbb{R}$, for all $x,y in mathbb{Q}$, there exists $w notin mathbb{Q}$ such that $x < w < y$. As well for all $q_1, q_2 notin mathbb{Q}$, there exists $r in mathbb{Q}$ such that $q_1 < r < q_2$
With this being the case there are two cases to examine:
$x in mathbb{Q}, a notin mathbb{Q}$ AND
$a in mathbb{Q}, x notin mathbb{Q}$
Without loss of generality let's suppose:
$x in mathbb{Q}, a notin mathbb{Q}$
The other follows directly.
Let $epsilon = 0$, $delta > 0$, and let $x = frac{delta}{2} + a$
Therefore by definition: $$ bigg| x - a bigg| = bigg| (frac{delta}{2} + a) - a bigg| = frac{delta}{2} < delta$$
This means: $$bigg|f(x) - f(a) bigg| = |1 - 0| = 1 > epsilon = 0$$
I feel that this is sufficient because by invoking the Archimedian properties I take care of the whole issue of the denseness with $mathbb{Q}$. But this is why I am asking the question here. Could I get feedback?
real-analysis proof-verification
asked Dec 3 '18 at 22:19
dc3rddc3rd
1,47011137
$endgroup$
add a comment |
$begingroup$
Let $S subset mathbb{R^n}$ and let $f$ be a function from $S$ into $mathbb{R^m}$. A function $f$ is continuous at $a$ if $forall epsilon > 0, exists delta > 0$ such that if $bigg| x - a bigg| < delta rightarrow bigg|f(x) - f(a) bigg| < epsilon$ For
Prove that the indicator function: $$f(x) = begin {cases} 0 & xnotin mathbb{Q}\ 1 & x in mathbb{Q} end{cases} $$
is discontinuous at every point $a in mathbb{R}$.
My professor assigned this exercise as practice for us to do. Specifically he asked us to show "formally" that it is the case that the indicator function is discontinuous. I understand exactly why it is discontinuous, I would like to know if the manner in which I proved it is "mathematically rigorous".
Proof:
By the Archimedian properties of $mathbb{R}$, for all $x,y in mathbb{Q}$, there exists $w notin mathbb{Q}$ such that $x < w < y$. As well for all $q_1, q_2 notin mathbb{Q}$, there exists $r in mathbb{Q}$ such that $q_1 < r < q_2$
With this being the case there are two cases to examine:
$x in mathbb{Q}, a notin mathbb{Q}$ AND
$a in mathbb{Q}, x notin mathbb{Q}$
Without loss of generality let's suppose:
$x in mathbb{Q}, a notin mathbb{Q}$
The other follows directly.
Let $epsilon = 0$, $delta > 0$, and let $x = frac{delta}{2} + a$
Therefore by definition: $$ bigg| x - a bigg| = bigg| (frac{delta}{2} + a) - a bigg| = frac{delta}{2} < delta$$
This means: $$bigg|f(x) - f(a) bigg| = |1 - 0| = 1 > epsilon = 0$$
I feel that this is sufficient because by invoking the Archimedian properties I take care of the whole issue of the denseness with $mathbb{Q}$. But this is why I am asking the question here. Could I get feedback?
real-analysis proof-verification
asked Dec 3 '18 at 22:19
dc3rddc3rd
1,47011137
$endgroup$
Let $S subset mathbb{R^n}$ and let $f$ be a function from $S$ into $mathbb{R^m}$. A function $f$ is continuous at $a$ if $forall epsilon > 0, exists delta > 0$ such that if $bigg| x - a bigg| < delta rightarrow bigg|f(x) - f(a) bigg| < epsilon$ For
Prove that the indicator function: $$f(x) = begin {cases} 0 & xnotin mathbb{Q}\ 1 & x in mathbb{Q} end{cases} $$
is discontinuous at every point $a in mathbb{R}$.
My professor assigned this exercise as practice for us to do. Specifically he asked us to show "formally" that it is the case that the indicator function is discontinuous. I understand exactly why it is discontinuous, I would like to know if the manner in which I proved it is "mathematically rigorous".
Proof:
By the Archimedian properties of $mathbb{R}$, for all $x,y in mathbb{Q}$, there exists $w notin mathbb{Q}$ such that $x < w < y$. As well for all $q_1, q_2 notin mathbb{Q}$, there exists $r in mathbb{Q}$ such that $q_1 < r < q_2$
With this being the case there are two cases to examine:
$x in mathbb{Q}, a notin mathbb{Q}$ AND
$a in mathbb{Q}, x notin mathbb{Q}$
Without loss of generality let's suppose:
$x in mathbb{Q}, a notin mathbb{Q}$
The other follows directly.
Let $epsilon = 0$, $delta > 0$, and let $x = frac{delta}{2} + a$
Therefore by definition: $$ bigg| x - a bigg| = bigg| (frac{delta}{2} + a) - a bigg| = frac{delta}{2} < delta$$
This means: $$bigg|f(x) - f(a) bigg| = |1 - 0| = 1 > epsilon = 0$$
I feel that this is sufficient because by invoking the Archimedian properties I take care of the whole issue of the denseness with $mathbb{Q}$. But this is why I am asking the question here. Could I get feedback?
real-analysis proof-verification
real-analysis proof-verification
asked Dec 3 '18 at 22:19
dc3rddc3rd
1,47011137
asked Dec 3 '18 at 22:19
dc3rddc3rd
1,47011137
asked Dec 3 '18 at 22:19
dc3rddc3rd
1,47011137
asked Dec 3 '18 at 22:19
dc3rddc3rd
1,47011137
asked Dec 3 '18 at 22:19
dc3rddc3rd
1,47011137
1,47011137
add a comment |
add a comment |
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