Show that $(x_n)_{ninmathbb{N}}$ converges to $x$ if and only if $(f(x_n))_{ninmathbb{N}}$ converges to...
$begingroup$
We are supposed to use a previously solved problem to help deduce this, but I can't figure out how they relate. The other problem goes as follows: "Suppose $g$ is a strictly increasing function and let $B$ be the range of $g$. Prove that the inverse of $g$ is continuous."
I managed to solve that part, but I don't know how to use it to solve the rest. Any help would be appreciated.
real-analysis limits analysis
edited Oct 31 '18 at 8:54
rtybase
11.1k21533
$endgroup$
add a comment |
$begingroup$
We are supposed to use a previously solved problem to help deduce this, but I can't figure out how they relate. The other problem goes as follows: "Suppose $g$ is a strictly increasing function and let $B$ be the range of $g$. Prove that the inverse of $g$ is continuous."
I managed to solve that part, but I don't know how to use it to solve the rest. Any help would be appreciated.
real-analysis limits analysis
edited Oct 31 '18 at 8:54
rtybase
11.1k21533
$endgroup$
$begingroup$
But this is false for a jump discontinuity. What am I missing?
$endgroup$
– Matt Samuel
Oct 30 '18 at 2:39
add a comment |
$begingroup$
We are supposed to use a previously solved problem to help deduce this, but I can't figure out how they relate. The other problem goes as follows: "Suppose $g$ is a strictly increasing function and let $B$ be the range of $g$. Prove that the inverse of $g$ is continuous."
I managed to solve that part, but I don't know how to use it to solve the rest. Any help would be appreciated.
real-analysis limits analysis
edited Oct 31 '18 at 8:54
rtybase
11.1k21533
$endgroup$
We are supposed to use a previously solved problem to help deduce this, but I can't figure out how they relate. The other problem goes as follows: "Suppose $g$ is a strictly increasing function and let $B$ be the range of $g$. Prove that the inverse of $g$ is continuous."
I managed to solve that part, but I don't know how to use it to solve the rest. Any help would be appreciated.
real-analysis limits analysis
real-analysis limits analysis
edited Oct 31 '18 at 8:54
rtybase
11.1k21533
edited Oct 31 '18 at 8:54
rtybase
11.1k21533
edited Oct 31 '18 at 8:54
rtybase
11.1k21533
edited Oct 31 '18 at 8:54
rtybase
11.1k21533
edited Oct 31 '18 at 8:54
rtybase
11.1k21533
11.1k21533
asked Oct 30 '18 at 2:05
user610107
$begingroup$
But this is false for a jump discontinuity. What am I missing?
$endgroup$
– Matt Samuel
Oct 30 '18 at 2:39
add a comment |
$begingroup$
But this is false for a jump discontinuity. What am I missing?
$endgroup$
– Matt Samuel
Oct 30 '18 at 2:39
$begingroup$
But this is false for a jump discontinuity. What am I missing?
$endgroup$
– Matt Samuel
Oct 30 '18 at 2:39
$begingroup$
But this is false for a jump discontinuity. What am I missing?
$endgroup$
– Matt Samuel
Oct 30 '18 at 2:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
After some more digging, and consulting some other sources, I think I have a sufficient answer.
For the reverse direction, we assume $f$ is continuous at $x$, and that $x_n$ approaches $x.$
Let $varepsilon>0.$
We need to find a $N in mathbb{N}$ such that $|f(x_n)-f(x)| <varepsilon$ for every $n geq N.$
Since $f$ is continuous at $x$, there exists a $delta >0$ such that $|f(y)-(fx)| <varepsilon$ whenever $|y-x|<delta.$
Since $x_n rightarrow x$, there exists $N in mathbb{N}$ such that $|x_n-x|< delta$ for every $n geq N$. We can use this $N$ in our definition above.
Therefore, $|f(x_n)-f(x)|< varepsilon$ for every $n geq N.$
For the forwards direction, we assume towards a contradiction that $f$ is not continuous at $x.$
Then, for some $varepsilon >0$, and for each $n geq N,$ there exists a $x_n$ such that $|x_n-x|<frac{1}{n}$, but $|f(x_n)-f(x)|geq varepsilon.$
This is a contradiction, because we are claiming that $x_n rightarrow x$ if and only if $f(x_n) rightarrow f(x).$
And that's the proof. I could be wrong, but to me it seems to make sense.
edited Dec 19 '18 at 5:57
Gaby Alfonso
1,006317
$endgroup$
add a comment |
$begingroup$
The question is false. The property you're describing is sequential continuity of $f$. This is equivalent to regular continuity of $f$. There exist functions that are strictly increasing but not continuous.
answered Oct 30 '18 at 4:08
Bailey WhitbreadBailey Whitbread
13
$endgroup$
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%2f2977011%2fshow-that-x-n-n-in-mathbbn-converges-to-x-if-and-only-if-fx-n-n%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
$begingroup$
After some more digging, and consulting some other sources, I think I have a sufficient answer.
For the reverse direction, we assume $f$ is continuous at $x$, and that $x_n$ approaches $x.$
Let $varepsilon>0.$
We need to find a $N in mathbb{N}$ such that $|f(x_n)-f(x)| <varepsilon$ for every $n geq N.$
Since $f$ is continuous at $x$, there exists a $delta >0$ such that $|f(y)-(fx)| <varepsilon$ whenever $|y-x|<delta.$
Since $x_n rightarrow x$, there exists $N in mathbb{N}$ such that $|x_n-x|< delta$ for every $n geq N$. We can use this $N$ in our definition above.
Therefore, $|f(x_n)-f(x)|< varepsilon$ for every $n geq N.$
For the forwards direction, we assume towards a contradiction that $f$ is not continuous at $x.$
Then, for some $varepsilon >0$, and for each $n geq N,$ there exists a $x_n$ such that $|x_n-x|<frac{1}{n}$, but $|f(x_n)-f(x)|geq varepsilon.$
This is a contradiction, because we are claiming that $x_n rightarrow x$ if and only if $f(x_n) rightarrow f(x).$
And that's the proof. I could be wrong, but to me it seems to make sense.
edited Dec 19 '18 at 5:57
Gaby Alfonso
1,006317
$endgroup$
add a comment |
$begingroup$
After some more digging, and consulting some other sources, I think I have a sufficient answer.
For the reverse direction, we assume $f$ is continuous at $x$, and that $x_n$ approaches $x.$
Let $varepsilon>0.$
We need to find a $N in mathbb{N}$ such that $|f(x_n)-f(x)| <varepsilon$ for every $n geq N.$
Since $f$ is continuous at $x$, there exists a $delta >0$ such that $|f(y)-(fx)| <varepsilon$ whenever $|y-x|<delta.$
Since $x_n rightarrow x$, there exists $N in mathbb{N}$ such that $|x_n-x|< delta$ for every $n geq N$. We can use this $N$ in our definition above.
Therefore, $|f(x_n)-f(x)|< varepsilon$ for every $n geq N.$
For the forwards direction, we assume towards a contradiction that $f$ is not continuous at $x.$
Then, for some $varepsilon >0$, and for each $n geq N,$ there exists a $x_n$ such that $|x_n-x|<frac{1}{n}$, but $|f(x_n)-f(x)|geq varepsilon.$
This is a contradiction, because we are claiming that $x_n rightarrow x$ if and only if $f(x_n) rightarrow f(x).$
And that's the proof. I could be wrong, but to me it seems to make sense.
edited Dec 19 '18 at 5:57
Gaby Alfonso
1,006317
$endgroup$
add a comment |
$begingroup$
After some more digging, and consulting some other sources, I think I have a sufficient answer.
For the reverse direction, we assume $f$ is continuous at $x$, and that $x_n$ approaches $x.$
Let $varepsilon>0.$
We need to find a $N in mathbb{N}$ such that $|f(x_n)-f(x)| <varepsilon$ for every $n geq N.$
Since $f$ is continuous at $x$, there exists a $delta >0$ such that $|f(y)-(fx)| <varepsilon$ whenever $|y-x|<delta.$
Since $x_n rightarrow x$, there exists $N in mathbb{N}$ such that $|x_n-x|< delta$ for every $n geq N$. We can use this $N$ in our definition above.
Therefore, $|f(x_n)-f(x)|< varepsilon$ for every $n geq N.$
For the forwards direction, we assume towards a contradiction that $f$ is not continuous at $x.$
Then, for some $varepsilon >0$, and for each $n geq N,$ there exists a $x_n$ such that $|x_n-x|<frac{1}{n}$, but $|f(x_n)-f(x)|geq varepsilon.$
This is a contradiction, because we are claiming that $x_n rightarrow x$ if and only if $f(x_n) rightarrow f(x).$
And that's the proof. I could be wrong, but to me it seems to make sense.
edited Dec 19 '18 at 5:57
Gaby Alfonso
1,006317
$endgroup$
After some more digging, and consulting some other sources, I think I have a sufficient answer.
For the reverse direction, we assume $f$ is continuous at $x$, and that $x_n$ approaches $x.$
Let $varepsilon>0.$
We need to find a $N in mathbb{N}$ such that $|f(x_n)-f(x)| <varepsilon$ for every $n geq N.$
Since $f$ is continuous at $x$, there exists a $delta >0$ such that $|f(y)-(fx)| <varepsilon$ whenever $|y-x|<delta.$
Since $x_n rightarrow x$, there exists $N in mathbb{N}$ such that $|x_n-x|< delta$ for every $n geq N$. We can use this $N$ in our definition above.
Therefore, $|f(x_n)-f(x)|< varepsilon$ for every $n geq N.$
For the forwards direction, we assume towards a contradiction that $f$ is not continuous at $x.$
Then, for some $varepsilon >0$, and for each $n geq N,$ there exists a $x_n$ such that $|x_n-x|<frac{1}{n}$, but $|f(x_n)-f(x)|geq varepsilon.$
This is a contradiction, because we are claiming that $x_n rightarrow x$ if and only if $f(x_n) rightarrow f(x).$
And that's the proof. I could be wrong, but to me it seems to make sense.
edited Dec 19 '18 at 5:57
Gaby Alfonso
1,006317
edited Dec 19 '18 at 5:57
Gaby Alfonso
1,006317
edited Dec 19 '18 at 5:57
Gaby Alfonso
1,006317
edited Dec 19 '18 at 5:57
Gaby Alfonso
1,006317
1,006317
answered Oct 31 '18 at 5:18
user610107
add a comment |
add a comment |
$begingroup$
The question is false. The property you're describing is sequential continuity of $f$. This is equivalent to regular continuity of $f$. There exist functions that are strictly increasing but not continuous.
answered Oct 30 '18 at 4:08
Bailey WhitbreadBailey Whitbread
13
$endgroup$
add a comment |
$begingroup$
The question is false. The property you're describing is sequential continuity of $f$. This is equivalent to regular continuity of $f$. There exist functions that are strictly increasing but not continuous.
answered Oct 30 '18 at 4:08
Bailey WhitbreadBailey Whitbread
13
$endgroup$
add a comment |
$begingroup$
The question is false. The property you're describing is sequential continuity of $f$. This is equivalent to regular continuity of $f$. There exist functions that are strictly increasing but not continuous.
answered Oct 30 '18 at 4:08
Bailey WhitbreadBailey Whitbread
13
$endgroup$
The question is false. The property you're describing is sequential continuity of $f$. This is equivalent to regular continuity of $f$. There exist functions that are strictly increasing but not continuous.
answered Oct 30 '18 at 4:08
Bailey WhitbreadBailey Whitbread
13
answered Oct 30 '18 at 4:08
Bailey WhitbreadBailey Whitbread
13
answered Oct 30 '18 at 4:08
Bailey WhitbreadBailey Whitbread
13
answered Oct 30 '18 at 4:08
Bailey WhitbreadBailey Whitbread
13
13
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%2f2977011%2fshow-that-x-n-n-in-mathbbn-converges-to-x-if-and-only-if-fx-n-n%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
$begingroup$
But this is false for a jump discontinuity. What am I missing?
$endgroup$
– Matt Samuel
Oct 30 '18 at 2:39