Continuity of functions of monotone sequences
$begingroup$
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 '18 at 13:25
user3482749
4,047818
asked Nov 26 '18 at 13:16
Aishwarya DeoreAishwarya Deore
324
$endgroup$
add a comment |
$begingroup$
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 '18 at 13:25
user3482749
4,047818
asked Nov 26 '18 at 13:16
Aishwarya DeoreAishwarya Deore
324
$endgroup$
$begingroup$
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.
$endgroup$
– Offlaw
Nov 26 '18 at 13:24
add a comment |
$begingroup$
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 '18 at 13:25
user3482749
4,047818
asked Nov 26 '18 at 13:16
Aishwarya DeoreAishwarya Deore
324
$endgroup$
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 '18 at 13:25
user3482749
4,047818
asked Nov 26 '18 at 13:16
Aishwarya DeoreAishwarya Deore
324
edited Nov 26 '18 at 13:25
user3482749
4,047818
asked Nov 26 '18 at 13:16
Aishwarya DeoreAishwarya Deore
324
edited Nov 26 '18 at 13:25
user3482749
4,047818
edited Nov 26 '18 at 13:25
user3482749
4,047818
edited Nov 26 '18 at 13:25
user3482749
4,047818
4,047818
asked Nov 26 '18 at 13:16
Aishwarya DeoreAishwarya Deore
324
asked Nov 26 '18 at 13:16
Aishwarya DeoreAishwarya Deore
324
asked Nov 26 '18 at 13:16
Aishwarya DeoreAishwarya Deore
324
324
$begingroup$
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.
$endgroup$
– Offlaw
Nov 26 '18 at 13:24
add a comment |
$begingroup$
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.
$endgroup$
– Offlaw
Nov 26 '18 at 13:24
$begingroup$
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.
$endgroup$
– Offlaw
Nov 26 '18 at 13:24
$begingroup$
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.
$endgroup$
– Offlaw
Nov 26 '18 at 13:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 '18 at 13:25
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
$begingroup$
“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.
$endgroup$
– Offlaw
Nov 26 '18 at 13:30
$begingroup$
"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
$endgroup$
– Aishwarya Deore
Nov 26 '18 at 13:42
$begingroup$
@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?
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:25
add a comment |
$begingroup$
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 '18 at 13:27
user3482749user3482749
4,047818
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014320%2fcontinuity-of-functions-of-monotone-sequences%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 '18 at 13:25
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
$begingroup$
“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.
$endgroup$
– Offlaw
Nov 26 '18 at 13:30
$begingroup$
"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
$endgroup$
– Aishwarya Deore
Nov 26 '18 at 13:42
$begingroup$
@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?
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:25
add a comment |
$begingroup$
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 '18 at 13:25
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
$begingroup$
“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.
$endgroup$
– Offlaw
Nov 26 '18 at 13:30
$begingroup$
"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
$endgroup$
– Aishwarya Deore
Nov 26 '18 at 13:42
$begingroup$
@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?
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:25
add a comment |
$begingroup$
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 '18 at 13:25
José Carlos SantosJosé Carlos Santos
156k22125227
$endgroup$
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 '18 at 13:25
José Carlos SantosJosé Carlos Santos
156k22125227
answered Nov 26 '18 at 13:25
José Carlos SantosJosé Carlos Santos
156k22125227
answered Nov 26 '18 at 13:25
José Carlos SantosJosé Carlos Santos
156k22125227
answered Nov 26 '18 at 13:25
José Carlos SantosJosé Carlos Santos
156k22125227
156k22125227
$begingroup$
“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.
$endgroup$
– Offlaw
Nov 26 '18 at 13:30
$begingroup$
"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
$endgroup$
– Aishwarya Deore
Nov 26 '18 at 13:42
$begingroup$
@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?
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:25
add a comment |
$begingroup$
“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.
$endgroup$
– Offlaw
Nov 26 '18 at 13:30
$begingroup$
"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
$endgroup$
– Aishwarya Deore
Nov 26 '18 at 13:42
$begingroup$
@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?
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:25
$begingroup$
“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.
$endgroup$
– Offlaw
Nov 26 '18 at 13:30
$begingroup$
“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.
$endgroup$
– Offlaw
Nov 26 '18 at 13:30
$begingroup$
"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
$endgroup$
– Aishwarya Deore
Nov 26 '18 at 13:42
$begingroup$
"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
$endgroup$
– Aishwarya Deore
Nov 26 '18 at 13:42
$begingroup$
@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?
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:25
$begingroup$
@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?
$endgroup$
– José Carlos Santos
Nov 26 '18 at 17:25
add a comment |
$begingroup$
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 '18 at 13:27
user3482749user3482749
4,047818
$endgroup$
add a comment |
$begingroup$
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 '18 at 13:27
user3482749user3482749
4,047818
$endgroup$
add a comment |
$begingroup$
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 '18 at 13:27
user3482749user3482749
4,047818
$endgroup$
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 '18 at 13:27
user3482749user3482749
4,047818
answered Nov 26 '18 at 13:27
user3482749user3482749
4,047818
answered Nov 26 '18 at 13:27
user3482749user3482749
4,047818
answered Nov 26 '18 at 13:27
user3482749user3482749
4,047818
4,047818
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014320%2fcontinuity-of-functions-of-monotone-sequences%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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.
$endgroup$
– Offlaw
Nov 26 '18 at 13:24