Continuity of functions of monotone sequences
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Suppose $lim_{ntoinfty} f(x_n) = f(c)$ for any monotone sequence $x_n$ approaching $c$. The prove that $f$ is continuous at $c$.
Solution: We prove it by contradiction. Assume $f$ is not continuous at $c$. Then there exists $varepsilon > 0$ such that for any $n$ belonging to $mathbb{N}$, there is $x_n$ such that $x_n$ approaches c but $|f(x_n) - f(c)| > varepsilon$. Then there is subsequence $x_{n_k}$ such that $lim_{ktoinfty} x_{n_k} = c$ and $x_{n_k}$ is monotone. Then by assumption we have $lim f(x_{n_k}) = f(c)$ which is a contradiction.
Is this right? Can someone explain why we have a contradiction?
real-analysis
edited Nov 26 at 13:25
user3482749
2,096414
asked Nov 26 at 13:16
Aishwarya Deore
324
add a comment |
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favorite
Suppose $lim_{ntoinfty} f(x_n) = f(c)$ for any monotone sequence $x_n$ approaching $c$. The prove that $f$ is continuous at $c$.
Solution: We prove it by contradiction. Assume $f$ is not continuous at $c$. Then there exists $varepsilon > 0$ such that for any $n$ belonging to $mathbb{N}$, there is $x_n$ such that $x_n$ approaches c but $|f(x_n) - f(c)| > varepsilon$. Then there is subsequence $x_{n_k}$ such that $lim_{ktoinfty} x_{n_k} = c$ and $x_{n_k}$ is monotone. Then by assumption we have $lim f(x_{n_k}) = f(c)$ which is a contradiction.
Is this right? Can someone explain why we have a contradiction?
real-analysis
edited Nov 26 at 13:25
user3482749
2,096414
asked Nov 26 at 13:16
Aishwarya Deore
324
If $f(x_n) to f(c)$ for any sequence ${x_n}$ converges to $c$, then also $f$ is continuous at $c$. Here only a special case is considered.
– Offlaw
Nov 26 at 13:24
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up vote
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down vote
favorite
Suppose $lim_{ntoinfty} f(x_n) = f(c)$ for any monotone sequence $x_n$ approaching $c$. The prove that $f$ is continuous at $c$.
Solution: We prove it by contradiction. Assume $f$ is not continuous at $c$. Then there exists $varepsilon > 0$ such that for any $n$ belonging to $mathbb{N}$, there is $x_n$ such that $x_n$ approaches c but $|f(x_n) - f(c)| > varepsilon$. Then there is subsequence $x_{n_k}$ such that $lim_{ktoinfty} x_{n_k} = c$ and $x_{n_k}$ is monotone. Then by assumption we have $lim f(x_{n_k}) = f(c)$ which is a contradiction.
Is this right? Can someone explain why we have a contradiction?
real-analysis
edited Nov 26 at 13:25
user3482749
2,096414
asked Nov 26 at 13:16
Aishwarya Deore
324
Suppose $lim_{ntoinfty} f(x_n) = f(c)$ for any monotone sequence $x_n$ approaching $c$. The prove that $f$ is continuous at $c$.
Solution: We prove it by contradiction. Assume $f$ is not continuous at $c$. Then there exists $varepsilon > 0$ such that for any $n$ belonging to $mathbb{N}$, there is $x_n$ such that $x_n$ approaches c but $|f(x_n) - f(c)| > varepsilon$. Then there is subsequence $x_{n_k}$ such that $lim_{ktoinfty} x_{n_k} = c$ and $x_{n_k}$ is monotone. Then by assumption we have $lim f(x_{n_k}) = f(c)$ which is a contradiction.
Is this right? Can someone explain why we have a contradiction?
real-analysis
real-analysis
edited Nov 26 at 13:25
user3482749
2,096414
asked Nov 26 at 13:16
Aishwarya Deore
324
edited Nov 26 at 13:25
user3482749
2,096414
asked Nov 26 at 13:16
Aishwarya Deore
324
edited Nov 26 at 13:25
user3482749
2,096414
edited Nov 26 at 13:25
user3482749
2,096414
edited Nov 26 at 13:25
user3482749
2,096414
2,096414
asked Nov 26 at 13:16
Aishwarya Deore
324
asked Nov 26 at 13:16
Aishwarya Deore
324
asked Nov 26 at 13:16
Aishwarya Deore
324
324
If $f(x_n) to f(c)$ for any sequence ${x_n}$ converges to $c$, then also $f$ is continuous at $c$. Here only a special case is considered.
– Offlaw
Nov 26 at 13:24
add a comment |
If $f(x_n) to f(c)$ for any sequence ${x_n}$ converges to $c$, then also $f$ is continuous at $c$. Here only a special case is considered.
– Offlaw
Nov 26 at 13:24
If $f(x_n) to f(c)$ for any sequence ${x_n}$ converges to $c$, then also $f$ is continuous at $c$. Here only a special case is considered.
– Offlaw
Nov 26 at 13:24
If $f(x_n) to f(c)$ for any sequence ${x_n}$ converges to $c$, then also $f$ is continuous at $c$. Here only a special case is considered.
– Offlaw
Nov 26 at 13:24
add a comment |
2 Answers
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No, it is not right. The assertion “there is $x_n$ such that $x_n$ approaches $c$ but $bigl|f(x_n)-f(c)bigr|>varepsilon$” makes no sense.
There is a $varepsilon>0$ such that, for each $delta>0$, there is a $xin(c-delta,c+delta)cap D_f$ such that $bigllvert f(x)-f(c)bigrrvertgeqslantvarepsilon$. In particular, for any natural $n$, there is a $x_ninleft(c-frac1n,c+frac1nright)cap D_f$ such that $bigllvert f(x_n)-f(c)bigrrvertgeqslantvarepsilon$. The sequence $(x_n)_{ninmathbb N}$ has a monotonic subsequence $(x_{n_k})_{kinmathbb N}$ and, since $lim_{ntoinfty}x_n=c$, $lim_{ktoinfty}x_{n_k}=c$. Therefore, we should have $lim_{ktoinfty}f(x_{n_k})=f(c)$. But we don't, since $(forall kinmathbb{N}):bigllvert f(x_{n_k})-f(c)bigrrvertgeqslantvarepsilon$. So, we have a contradiction here.
answered Nov 26 at 13:25
José Carlos Santos
146k22116216
“there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>epsilon$" - Why does this not make sense? Please I can't understand.
– Offlaw
Nov 26 at 13:30
"x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε
– Aishwarya Deore
Nov 26 at 13:42
@AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $mathbb N$, there is $x_n$” mean?
– José Carlos Santos
Nov 26 at 17:25
add a comment |
up vote
0
down vote
No. Mostly, your third sentence is mangled. Here's a corrected version:
If $f$ is not left continuous at $c$, then there is some $varepsilon > 0$ such that for any $delta > 0$, there is some $x in (c-delta,c)$ such that $|f(x)-f(c)| geq varepsilon$. Choosing, in particular, for each $n in mathbb{N}$, $delta = frac{1}{n}$, and choosing some $x_n in (c-delta,c)$, we obtain a sequence $(x_n)$ such that $(x_n)to c$ but $|f(x_n)-f(c)| geq varepsilon$ for all $n$. Now, every sequence has a monotone subsequence, so in particular $(x_n)$ has a monotone subsequence $(x_{n_k})$, and $(x_{n_k})to c$, since it's a subsequence of $(x_n)$, and $|f(x_{n_k})-f(c)|geqvarepsilon$ for all $k$. This contradicts our hypothesis about $f$, so $f$ is left continuous.
An identical proof (with $(c-delta,c)$ replaced by $(c,c+delta)$) shows that $f$ is right-continuous [or you could just patch this into the main proof if you prefer], so $f$ is both left- and right-continuous, so is continuous.
answered Nov 26 at 13:27
user3482749
2,096414
add a comment |
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2 Answers
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2 Answers
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up vote
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No, it is not right. The assertion “there is $x_n$ such that $x_n$ approaches $c$ but $bigl|f(x_n)-f(c)bigr|>varepsilon$” makes no sense.
There is a $varepsilon>0$ such that, for each $delta>0$, there is a $xin(c-delta,c+delta)cap D_f$ such that $bigllvert f(x)-f(c)bigrrvertgeqslantvarepsilon$. In particular, for any natural $n$, there is a $x_ninleft(c-frac1n,c+frac1nright)cap D_f$ such that $bigllvert f(x_n)-f(c)bigrrvertgeqslantvarepsilon$. The sequence $(x_n)_{ninmathbb N}$ has a monotonic subsequence $(x_{n_k})_{kinmathbb N}$ and, since $lim_{ntoinfty}x_n=c$, $lim_{ktoinfty}x_{n_k}=c$. Therefore, we should have $lim_{ktoinfty}f(x_{n_k})=f(c)$. But we don't, since $(forall kinmathbb{N}):bigllvert f(x_{n_k})-f(c)bigrrvertgeqslantvarepsilon$. So, we have a contradiction here.
answered Nov 26 at 13:25
José Carlos Santos
146k22116216
“there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>epsilon$" - Why does this not make sense? Please I can't understand.
– Offlaw
Nov 26 at 13:30
"x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε
– Aishwarya Deore
Nov 26 at 13:42
@AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $mathbb N$, there is $x_n$” mean?
– José Carlos Santos
Nov 26 at 17:25
add a comment |
up vote
0
down vote
No, it is not right. The assertion “there is $x_n$ such that $x_n$ approaches $c$ but $bigl|f(x_n)-f(c)bigr|>varepsilon$” makes no sense.
There is a $varepsilon>0$ such that, for each $delta>0$, there is a $xin(c-delta,c+delta)cap D_f$ such that $bigllvert f(x)-f(c)bigrrvertgeqslantvarepsilon$. In particular, for any natural $n$, there is a $x_ninleft(c-frac1n,c+frac1nright)cap D_f$ such that $bigllvert f(x_n)-f(c)bigrrvertgeqslantvarepsilon$. The sequence $(x_n)_{ninmathbb N}$ has a monotonic subsequence $(x_{n_k})_{kinmathbb N}$ and, since $lim_{ntoinfty}x_n=c$, $lim_{ktoinfty}x_{n_k}=c$. Therefore, we should have $lim_{ktoinfty}f(x_{n_k})=f(c)$. But we don't, since $(forall kinmathbb{N}):bigllvert f(x_{n_k})-f(c)bigrrvertgeqslantvarepsilon$. So, we have a contradiction here.
answered Nov 26 at 13:25
José Carlos Santos
146k22116216
“there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>epsilon$" - Why does this not make sense? Please I can't understand.
– Offlaw
Nov 26 at 13:30
"x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε
– Aishwarya Deore
Nov 26 at 13:42
@AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $mathbb N$, there is $x_n$” mean?
– José Carlos Santos
Nov 26 at 17:25
add a comment |
up vote
0
down vote
up vote
0
down vote
No, it is not right. The assertion “there is $x_n$ such that $x_n$ approaches $c$ but $bigl|f(x_n)-f(c)bigr|>varepsilon$” makes no sense.
There is a $varepsilon>0$ such that, for each $delta>0$, there is a $xin(c-delta,c+delta)cap D_f$ such that $bigllvert f(x)-f(c)bigrrvertgeqslantvarepsilon$. In particular, for any natural $n$, there is a $x_ninleft(c-frac1n,c+frac1nright)cap D_f$ such that $bigllvert f(x_n)-f(c)bigrrvertgeqslantvarepsilon$. The sequence $(x_n)_{ninmathbb N}$ has a monotonic subsequence $(x_{n_k})_{kinmathbb N}$ and, since $lim_{ntoinfty}x_n=c$, $lim_{ktoinfty}x_{n_k}=c$. Therefore, we should have $lim_{ktoinfty}f(x_{n_k})=f(c)$. But we don't, since $(forall kinmathbb{N}):bigllvert f(x_{n_k})-f(c)bigrrvertgeqslantvarepsilon$. So, we have a contradiction here.
answered Nov 26 at 13:25
José Carlos Santos
146k22116216
No, it is not right. The assertion “there is $x_n$ such that $x_n$ approaches $c$ but $bigl|f(x_n)-f(c)bigr|>varepsilon$” makes no sense.
There is a $varepsilon>0$ such that, for each $delta>0$, there is a $xin(c-delta,c+delta)cap D_f$ such that $bigllvert f(x)-f(c)bigrrvertgeqslantvarepsilon$. In particular, for any natural $n$, there is a $x_ninleft(c-frac1n,c+frac1nright)cap D_f$ such that $bigllvert f(x_n)-f(c)bigrrvertgeqslantvarepsilon$. The sequence $(x_n)_{ninmathbb N}$ has a monotonic subsequence $(x_{n_k})_{kinmathbb N}$ and, since $lim_{ntoinfty}x_n=c$, $lim_{ktoinfty}x_{n_k}=c$. Therefore, we should have $lim_{ktoinfty}f(x_{n_k})=f(c)$. But we don't, since $(forall kinmathbb{N}):bigllvert f(x_{n_k})-f(c)bigrrvertgeqslantvarepsilon$. So, we have a contradiction here.
answered Nov 26 at 13:25
José Carlos Santos
146k22116216
answered Nov 26 at 13:25
José Carlos Santos
146k22116216
answered Nov 26 at 13:25
José Carlos Santos
146k22116216
answered Nov 26 at 13:25
José Carlos Santos
146k22116216
146k22116216
“there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>epsilon$" - Why does this not make sense? Please I can't understand.
– Offlaw
Nov 26 at 13:30
"x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε
– Aishwarya Deore
Nov 26 at 13:42
@AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $mathbb N$, there is $x_n$” mean?
– José Carlos Santos
Nov 26 at 17:25
add a comment |
“there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>epsilon$" - Why does this not make sense? Please I can't understand.
– Offlaw
Nov 26 at 13:30
"x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε
– Aishwarya Deore
Nov 26 at 13:42
@AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $mathbb N$, there is $x_n$” mean?
– José Carlos Santos
Nov 26 at 17:25
“there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>epsilon$" - Why does this not make sense? Please I can't understand.
– Offlaw
Nov 26 at 13:30
“there is $x_n$ such that $x_n$ approaches $c$ but $∣f(x_n)−f(c)∣>epsilon$" - Why does this not make sense? Please I can't understand.
– Offlaw
Nov 26 at 13:30
"x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε
– Aishwarya Deore
Nov 26 at 13:42
"x∈(c−δ,c+δ)∩Df such that ∣f(x)−f(c)∣⩾ε. In particular, for any natural n, there is a xn∈(c−1n,c+1n)∩Df such that ∣f(xn)−f(c)∣⩾ε. .... Isn't this exactly what I have written? choosing delta as i/n we say that there is xn such that xn is arbitarily close to c but ∣f(xn)−f(c)∣⩾ε
– Aishwarya Deore
Nov 26 at 13:42
@AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $mathbb N$, there is $x_n$” mean?
– José Carlos Santos
Nov 26 at 17:25
@AishwaryaDeore I disagree. To say that there is a (number) $n$ such that $x_n$ approaches $c$ makes no sense. A number doesn't approach anything. It just stays there. And if you had a sequence in mind, not just a number, then what does “for any $n$ belonging to $mathbb N$, there is $x_n$” mean?
– José Carlos Santos
Nov 26 at 17:25
add a comment |
up vote
0
down vote
No. Mostly, your third sentence is mangled. Here's a corrected version:
If $f$ is not left continuous at $c$, then there is some $varepsilon > 0$ such that for any $delta > 0$, there is some $x in (c-delta,c)$ such that $|f(x)-f(c)| geq varepsilon$. Choosing, in particular, for each $n in mathbb{N}$, $delta = frac{1}{n}$, and choosing some $x_n in (c-delta,c)$, we obtain a sequence $(x_n)$ such that $(x_n)to c$ but $|f(x_n)-f(c)| geq varepsilon$ for all $n$. Now, every sequence has a monotone subsequence, so in particular $(x_n)$ has a monotone subsequence $(x_{n_k})$, and $(x_{n_k})to c$, since it's a subsequence of $(x_n)$, and $|f(x_{n_k})-f(c)|geqvarepsilon$ for all $k$. This contradicts our hypothesis about $f$, so $f$ is left continuous.
An identical proof (with $(c-delta,c)$ replaced by $(c,c+delta)$) shows that $f$ is right-continuous [or you could just patch this into the main proof if you prefer], so $f$ is both left- and right-continuous, so is continuous.
answered Nov 26 at 13:27
user3482749
2,096414
add a comment |
up vote
0
down vote
No. Mostly, your third sentence is mangled. Here's a corrected version:
If $f$ is not left continuous at $c$, then there is some $varepsilon > 0$ such that for any $delta > 0$, there is some $x in (c-delta,c)$ such that $|f(x)-f(c)| geq varepsilon$. Choosing, in particular, for each $n in mathbb{N}$, $delta = frac{1}{n}$, and choosing some $x_n in (c-delta,c)$, we obtain a sequence $(x_n)$ such that $(x_n)to c$ but $|f(x_n)-f(c)| geq varepsilon$ for all $n$. Now, every sequence has a monotone subsequence, so in particular $(x_n)$ has a monotone subsequence $(x_{n_k})$, and $(x_{n_k})to c$, since it's a subsequence of $(x_n)$, and $|f(x_{n_k})-f(c)|geqvarepsilon$ for all $k$. This contradicts our hypothesis about $f$, so $f$ is left continuous.
An identical proof (with $(c-delta,c)$ replaced by $(c,c+delta)$) shows that $f$ is right-continuous [or you could just patch this into the main proof if you prefer], so $f$ is both left- and right-continuous, so is continuous.
answered Nov 26 at 13:27
user3482749
2,096414
add a comment |
up vote
0
down vote
up vote
0
down vote
No. Mostly, your third sentence is mangled. Here's a corrected version:
If $f$ is not left continuous at $c$, then there is some $varepsilon > 0$ such that for any $delta > 0$, there is some $x in (c-delta,c)$ such that $|f(x)-f(c)| geq varepsilon$. Choosing, in particular, for each $n in mathbb{N}$, $delta = frac{1}{n}$, and choosing some $x_n in (c-delta,c)$, we obtain a sequence $(x_n)$ such that $(x_n)to c$ but $|f(x_n)-f(c)| geq varepsilon$ for all $n$. Now, every sequence has a monotone subsequence, so in particular $(x_n)$ has a monotone subsequence $(x_{n_k})$, and $(x_{n_k})to c$, since it's a subsequence of $(x_n)$, and $|f(x_{n_k})-f(c)|geqvarepsilon$ for all $k$. This contradicts our hypothesis about $f$, so $f$ is left continuous.
An identical proof (with $(c-delta,c)$ replaced by $(c,c+delta)$) shows that $f$ is right-continuous [or you could just patch this into the main proof if you prefer], so $f$ is both left- and right-continuous, so is continuous.
answered Nov 26 at 13:27
user3482749
2,096414
No. Mostly, your third sentence is mangled. Here's a corrected version:
If $f$ is not left continuous at $c$, then there is some $varepsilon > 0$ such that for any $delta > 0$, there is some $x in (c-delta,c)$ such that $|f(x)-f(c)| geq varepsilon$. Choosing, in particular, for each $n in mathbb{N}$, $delta = frac{1}{n}$, and choosing some $x_n in (c-delta,c)$, we obtain a sequence $(x_n)$ such that $(x_n)to c$ but $|f(x_n)-f(c)| geq varepsilon$ for all $n$. Now, every sequence has a monotone subsequence, so in particular $(x_n)$ has a monotone subsequence $(x_{n_k})$, and $(x_{n_k})to c$, since it's a subsequence of $(x_n)$, and $|f(x_{n_k})-f(c)|geqvarepsilon$ for all $k$. This contradicts our hypothesis about $f$, so $f$ is left continuous.
An identical proof (with $(c-delta,c)$ replaced by $(c,c+delta)$) shows that $f$ is right-continuous [or you could just patch this into the main proof if you prefer], so $f$ is both left- and right-continuous, so is continuous.
answered Nov 26 at 13:27
user3482749
2,096414
answered Nov 26 at 13:27
user3482749
2,096414
answered Nov 26 at 13:27
user3482749
2,096414
answered Nov 26 at 13:27
user3482749
2,096414
2,096414
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If $f(x_n) to f(c)$ for any sequence ${x_n}$ converges to $c$, then also $f$ is continuous at $c$. Here only a special case is considered.
– Offlaw
Nov 26 at 13:24