Question about a function being continuousA problem on summing real numbers all taken to the same exponentProving a function is continuous on all irrational numbersHow can I prove this function is not continuous for every point other than 0?At what points is this piecewise function continuous?Explain about proofcontinuous but not absolutely continuous functionProb. 17, Chap. 4 in Baby Rudin: A real function can have at most countably many simple discontinuities on a given segmentGiven two real numbers $a$ and $b$ such that $a<b$, what about the convergence of these two sequences?Understanding a functionAlmost everywhere differentiability of a function
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Question about a function being continuous
A problem on summing real numbers all taken to the same exponentProving a function is continuous on all irrational numbersHow can I prove this function is not continuous for every point other than 0?At what points is this piecewise function continuous?Explain about proofcontinuous but not absolutely continuous functionProb. 17, Chap. 4 in Baby Rudin: A real function can have at most countably many simple discontinuities on a given segmentGiven two real numbers $a$ and $b$ such that $a<b$, what about the convergence of these two sequences?Understanding a functionAlmost everywhere differentiability of a function
$begingroup$
Let $r_1,r_2,...$ be the set of rational numbers in $[0,1]$ and
$g_n(x)=begincasesfrac1sqrt & xneq r_n; \ 3 & x=r_n.endcases$
This is continuous by definition ?
for all $ϵ>0$, there exists $δ>0$ such that for all $y$, if $|x−y|<δ$, then $|f(x)−f(y)|<ϵ$.
analysis
edited Mar 29 at 10:49
J. W. Tanner
4,4691320
asked Mar 29 at 10:29
kim wenasakim wenasa
194
$endgroup$
add a comment |
$begingroup$
Let $r_1,r_2,...$ be the set of rational numbers in $[0,1]$ and
$g_n(x)=begincasesfrac1sqrt & xneq r_n; \ 3 & x=r_n.endcases$
This is continuous by definition ?
for all $ϵ>0$, there exists $δ>0$ such that for all $y$, if $|x−y|<δ$, then $|f(x)−f(y)|<ϵ$.
analysis
edited Mar 29 at 10:49
J. W. Tanner
4,4691320
asked Mar 29 at 10:29
kim wenasakim wenasa
194
$endgroup$
add a comment |
$begingroup$
Let $r_1,r_2,...$ be the set of rational numbers in $[0,1]$ and
$g_n(x)=begincasesfrac1sqrt & xneq r_n; \ 3 & x=r_n.endcases$
This is continuous by definition ?
for all $ϵ>0$, there exists $δ>0$ such that for all $y$, if $|x−y|<δ$, then $|f(x)−f(y)|<ϵ$.
analysis
edited Mar 29 at 10:49
J. W. Tanner
4,4691320
asked Mar 29 at 10:29
kim wenasakim wenasa
194
$endgroup$
Let $r_1,r_2,...$ be the set of rational numbers in $[0,1]$ and
$g_n(x)=begincasesfrac1sqrt & xneq r_n; \ 3 & x=r_n.endcases$
This is continuous by definition ?
for all $ϵ>0$, there exists $δ>0$ such that for all $y$, if $|x−y|<δ$, then $|f(x)−f(y)|<ϵ$.
analysis
analysis
edited Mar 29 at 10:49
J. W. Tanner
4,4691320
asked Mar 29 at 10:29
kim wenasakim wenasa
194
edited Mar 29 at 10:49
J. W. Tanner
4,4691320
asked Mar 29 at 10:29
kim wenasakim wenasa
194
edited Mar 29 at 10:49
J. W. Tanner
4,4691320
edited Mar 29 at 10:49
J. W. Tanner
4,4691320
edited Mar 29 at 10:49
J. W. Tanner
4,4691320
4,4691320
asked Mar 29 at 10:29
kim wenasakim wenasa
194
asked Mar 29 at 10:29
kim wenasakim wenasa
194
asked Mar 29 at 10:29
kim wenasakim wenasa
194
194
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$g_n$ is not continuous at $r_n$: if $(x_k)$ is a sequence in $[0,1] setminus r_n$ with $x_k to r_n$ as $k to infty$, then $g_n(x_k) to infty$ as $k to infty.$
answered Mar 29 at 10:39
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
It's discontinuous, for $x$ close to $r_n$ values of $frac1sqrt$ get arbitrarily large, so for small $varepsilon$ you won't find such $delta$.
answered Mar 29 at 10:39
Maja BlumensteinMaja Blumenstein
1096
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
$g_n$ is not continuous at $r_n$: if $(x_k)$ is a sequence in $[0,1] setminus r_n$ with $x_k to r_n$ as $k to infty$, then $g_n(x_k) to infty$ as $k to infty.$
answered Mar 29 at 10:39
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
$g_n$ is not continuous at $r_n$: if $(x_k)$ is a sequence in $[0,1] setminus r_n$ with $x_k to r_n$ as $k to infty$, then $g_n(x_k) to infty$ as $k to infty.$
answered Mar 29 at 10:39
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
$g_n$ is not continuous at $r_n$: if $(x_k)$ is a sequence in $[0,1] setminus r_n$ with $x_k to r_n$ as $k to infty$, then $g_n(x_k) to infty$ as $k to infty.$
answered Mar 29 at 10:39
FredFred
48.6k11849
$endgroup$
$g_n$ is not continuous at $r_n$: if $(x_k)$ is a sequence in $[0,1] setminus r_n$ with $x_k to r_n$ as $k to infty$, then $g_n(x_k) to infty$ as $k to infty.$
answered Mar 29 at 10:39
FredFred
48.6k11849
answered Mar 29 at 10:39
FredFred
48.6k11849
answered Mar 29 at 10:39
FredFred
48.6k11849
answered Mar 29 at 10:39
FredFred
48.6k11849
48.6k11849
add a comment |
add a comment |
$begingroup$
It's discontinuous, for $x$ close to $r_n$ values of $frac1sqrt$ get arbitrarily large, so for small $varepsilon$ you won't find such $delta$.
answered Mar 29 at 10:39
Maja BlumensteinMaja Blumenstein
1096
$endgroup$
add a comment |
$begingroup$
It's discontinuous, for $x$ close to $r_n$ values of $frac1sqrt$ get arbitrarily large, so for small $varepsilon$ you won't find such $delta$.
answered Mar 29 at 10:39
Maja BlumensteinMaja Blumenstein
1096
$endgroup$
add a comment |
$begingroup$
It's discontinuous, for $x$ close to $r_n$ values of $frac1sqrt$ get arbitrarily large, so for small $varepsilon$ you won't find such $delta$.
answered Mar 29 at 10:39
Maja BlumensteinMaja Blumenstein
1096
$endgroup$
It's discontinuous, for $x$ close to $r_n$ values of $frac1sqrt$ get arbitrarily large, so for small $varepsilon$ you won't find such $delta$.
answered Mar 29 at 10:39
Maja BlumensteinMaja Blumenstein
1096
answered Mar 29 at 10:39
Maja BlumensteinMaja Blumenstein
1096
answered Mar 29 at 10:39
Maja BlumensteinMaja Blumenstein
1096
answered Mar 29 at 10:39
Maja BlumensteinMaja Blumenstein
1096
1096
add a comment |
add a comment |
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