Continuity of function $f(x)$ at irrational numbers.
$begingroup$
Let ${q_k: k in mathbb N^+}$ be an enumeration of the rational numbers in $[0, 1)$ and let $(a_k)$ be a sequence of strictly positive real numbers such that
$$
sum_{k=1}^infty a_k = 1
$$
Denote $S(x) = {k in mathbb N^+: q_k in [0, x)}$ i.e. indices of rational numbers in $[0, x)$ and define $f:[0, 1] rightarrow mathbb R$ by
$$
f(x) = begin{cases}
sum_{k in S(x)} a_k & (x > 0) \
0 & (x = 0)
end{cases}
$$
I know that $f(x)$ is discontinuous at every positive rational number. I speculate that $f(x)$ is continuous at irrational numbers, but I have no idea how to prove or disprove it.
real-analysis calculus continuity
asked Dec 7 '18 at 10:16
ritemeriteme
853
$endgroup$
add a comment |
$begingroup$
Let ${q_k: k in mathbb N^+}$ be an enumeration of the rational numbers in $[0, 1)$ and let $(a_k)$ be a sequence of strictly positive real numbers such that
$$
sum_{k=1}^infty a_k = 1
$$
Denote $S(x) = {k in mathbb N^+: q_k in [0, x)}$ i.e. indices of rational numbers in $[0, x)$ and define $f:[0, 1] rightarrow mathbb R$ by
$$
f(x) = begin{cases}
sum_{k in S(x)} a_k & (x > 0) \
0 & (x = 0)
end{cases}
$$
I know that $f(x)$ is discontinuous at every positive rational number. I speculate that $f(x)$ is continuous at irrational numbers, but I have no idea how to prove or disprove it.
real-analysis calculus continuity
asked Dec 7 '18 at 10:16
ritemeriteme
853
$endgroup$
add a comment |
$begingroup$
Let ${q_k: k in mathbb N^+}$ be an enumeration of the rational numbers in $[0, 1)$ and let $(a_k)$ be a sequence of strictly positive real numbers such that
$$
sum_{k=1}^infty a_k = 1
$$
Denote $S(x) = {k in mathbb N^+: q_k in [0, x)}$ i.e. indices of rational numbers in $[0, x)$ and define $f:[0, 1] rightarrow mathbb R$ by
$$
f(x) = begin{cases}
sum_{k in S(x)} a_k & (x > 0) \
0 & (x = 0)
end{cases}
$$
I know that $f(x)$ is discontinuous at every positive rational number. I speculate that $f(x)$ is continuous at irrational numbers, but I have no idea how to prove or disprove it.
real-analysis calculus continuity
asked Dec 7 '18 at 10:16
ritemeriteme
853
$endgroup$
Let ${q_k: k in mathbb N^+}$ be an enumeration of the rational numbers in $[0, 1)$ and let $(a_k)$ be a sequence of strictly positive real numbers such that
$$
sum_{k=1}^infty a_k = 1
$$
Denote $S(x) = {k in mathbb N^+: q_k in [0, x)}$ i.e. indices of rational numbers in $[0, x)$ and define $f:[0, 1] rightarrow mathbb R$ by
$$
f(x) = begin{cases}
sum_{k in S(x)} a_k & (x > 0) \
0 & (x = 0)
end{cases}
$$
I know that $f(x)$ is discontinuous at every positive rational number. I speculate that $f(x)$ is continuous at irrational numbers, but I have no idea how to prove or disprove it.
real-analysis calculus continuity
real-analysis calculus continuity
asked Dec 7 '18 at 10:16
ritemeriteme
853
asked Dec 7 '18 at 10:16
ritemeriteme
853
asked Dec 7 '18 at 10:16
ritemeriteme
853
asked Dec 7 '18 at 10:16
ritemeriteme
853
asked Dec 7 '18 at 10:16
ritemeriteme
853
853
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can find an answer in this thread. Although the situation is more special, because there $a_k = 2^{-k}$ is chosen, the proof doesn't change. Let $x in mathbb{R}$ be irrational, and let $N in mathbb{N}$ with $sum_{k=N1}^infty a_n < varepsilon/2$. Now define
$$delta:= min_{i=1,ldots,N}|q_i-x|.$$
For any $|y-x| < delta$ we have $q_k in [0,x)$ if and only if $q_k in [0,y)$ for all $k=1,ldots,N$. Thus $$S(x) cap [1,N] = S(y) cap [1,N]$$
and therefore
$$|f(x)-f(y)| le 2 sum_{k=N+1}^infty a_k<varepsilon.$$
answered Dec 7 '18 at 12:19
p4schp4sch
5,430318
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can find an answer in this thread. Although the situation is more special, because there $a_k = 2^{-k}$ is chosen, the proof doesn't change. Let $x in mathbb{R}$ be irrational, and let $N in mathbb{N}$ with $sum_{k=N1}^infty a_n < varepsilon/2$. Now define
$$delta:= min_{i=1,ldots,N}|q_i-x|.$$
For any $|y-x| < delta$ we have $q_k in [0,x)$ if and only if $q_k in [0,y)$ for all $k=1,ldots,N$. Thus $$S(x) cap [1,N] = S(y) cap [1,N]$$
and therefore
$$|f(x)-f(y)| le 2 sum_{k=N+1}^infty a_k<varepsilon.$$
answered Dec 7 '18 at 12:19
p4schp4sch
5,430318
$endgroup$
add a comment |
$begingroup$
You can find an answer in this thread. Although the situation is more special, because there $a_k = 2^{-k}$ is chosen, the proof doesn't change. Let $x in mathbb{R}$ be irrational, and let $N in mathbb{N}$ with $sum_{k=N1}^infty a_n < varepsilon/2$. Now define
$$delta:= min_{i=1,ldots,N}|q_i-x|.$$
For any $|y-x| < delta$ we have $q_k in [0,x)$ if and only if $q_k in [0,y)$ for all $k=1,ldots,N$. Thus $$S(x) cap [1,N] = S(y) cap [1,N]$$
and therefore
$$|f(x)-f(y)| le 2 sum_{k=N+1}^infty a_k<varepsilon.$$
answered Dec 7 '18 at 12:19
p4schp4sch
5,430318
$endgroup$
add a comment |
$begingroup$
You can find an answer in this thread. Although the situation is more special, because there $a_k = 2^{-k}$ is chosen, the proof doesn't change. Let $x in mathbb{R}$ be irrational, and let $N in mathbb{N}$ with $sum_{k=N1}^infty a_n < varepsilon/2$. Now define
$$delta:= min_{i=1,ldots,N}|q_i-x|.$$
For any $|y-x| < delta$ we have $q_k in [0,x)$ if and only if $q_k in [0,y)$ for all $k=1,ldots,N$. Thus $$S(x) cap [1,N] = S(y) cap [1,N]$$
and therefore
$$|f(x)-f(y)| le 2 sum_{k=N+1}^infty a_k<varepsilon.$$
answered Dec 7 '18 at 12:19
p4schp4sch
5,430318
$endgroup$
You can find an answer in this thread. Although the situation is more special, because there $a_k = 2^{-k}$ is chosen, the proof doesn't change. Let $x in mathbb{R}$ be irrational, and let $N in mathbb{N}$ with $sum_{k=N1}^infty a_n < varepsilon/2$. Now define
$$delta:= min_{i=1,ldots,N}|q_i-x|.$$
For any $|y-x| < delta$ we have $q_k in [0,x)$ if and only if $q_k in [0,y)$ for all $k=1,ldots,N$. Thus $$S(x) cap [1,N] = S(y) cap [1,N]$$
and therefore
$$|f(x)-f(y)| le 2 sum_{k=N+1}^infty a_k<varepsilon.$$
answered Dec 7 '18 at 12:19
p4schp4sch
5,430318
answered Dec 7 '18 at 12:19
p4schp4sch
5,430318
answered Dec 7 '18 at 12:19
p4schp4sch
5,430318
answered Dec 7 '18 at 12:19
p4schp4sch
5,430318
5,430318
add a comment |
add a comment |
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