$f(x)= (f_1(x),f_2(x))$ is continuous on X $iff $ $f_1$ and $f_2$ are continuous
$begingroup$
Let $X$ be any metric space. Let $f_1:Xto R$ and $F_2:Xto R $ be
defined. Then
$f:Xto R^2$
$f(x)= (f_1(x),f_2(x))$ is continuous on X
$iff $
$f_1$ and $f_2$ are continuous.
Instead of proving this via $epsilon -delta$ definition, can't we straight away use the sequential criterion to prove this?
$f:Xto R^2$
$f(x)= (f_1(x),f_2(x))$ is continuous on X
$iff$
$forall langle x_n rangle in X $ such that $langle x_n rangle to ain X$
we have $f(x_n)= (f_1(x_n),f_2(x_n)) to (f_1(a),f_2(a)) ; ; forall ain X$
$iff$
$f_1(x_n) to f1(a)$ and $f_2(x_n) to f_2(a)$
$iff$
Both $f_1$ and $f_2 $ are continuous on X.
general-topology continuity metric-spaces
$endgroup$
add a comment |
$begingroup$
Let $X$ be any metric space. Let $f_1:Xto R$ and $F_2:Xto R $ be
defined. Then
$f:Xto R^2$
$f(x)= (f_1(x),f_2(x))$ is continuous on X
$iff $
$f_1$ and $f_2$ are continuous.
Instead of proving this via $epsilon -delta$ definition, can't we straight away use the sequential criterion to prove this?
$f:Xto R^2$
$f(x)= (f_1(x),f_2(x))$ is continuous on X
$iff$
$forall langle x_n rangle in X $ such that $langle x_n rangle to ain X$
we have $f(x_n)= (f_1(x_n),f_2(x_n)) to (f_1(a),f_2(a)) ; ; forall ain X$
$iff$
$f_1(x_n) to f1(a)$ and $f_2(x_n) to f_2(a)$
$iff$
Both $f_1$ and $f_2 $ are continuous on X.
general-topology continuity metric-spaces
$endgroup$
$begingroup$
You proof is surely valid.
$endgroup$
– Kavi Rama Murthy
Nov 25 '18 at 5:21
$begingroup$
you can also consider messing about with inclusions and projections, again to avoid writing much anything down
$endgroup$
– qbert
Nov 25 '18 at 5:24
add a comment |
$begingroup$
Let $X$ be any metric space. Let $f_1:Xto R$ and $F_2:Xto R $ be
defined. Then
$f:Xto R^2$
$f(x)= (f_1(x),f_2(x))$ is continuous on X
$iff $
$f_1$ and $f_2$ are continuous.
Instead of proving this via $epsilon -delta$ definition, can't we straight away use the sequential criterion to prove this?
$f:Xto R^2$
$f(x)= (f_1(x),f_2(x))$ is continuous on X
$iff$
$forall langle x_n rangle in X $ such that $langle x_n rangle to ain X$
we have $f(x_n)= (f_1(x_n),f_2(x_n)) to (f_1(a),f_2(a)) ; ; forall ain X$
$iff$
$f_1(x_n) to f1(a)$ and $f_2(x_n) to f_2(a)$
$iff$
Both $f_1$ and $f_2 $ are continuous on X.
general-topology continuity metric-spaces
$endgroup$
Let $X$ be any metric space. Let $f_1:Xto R$ and $F_2:Xto R $ be
defined. Then
$f:Xto R^2$
$f(x)= (f_1(x),f_2(x))$ is continuous on X
$iff $
$f_1$ and $f_2$ are continuous.
Instead of proving this via $epsilon -delta$ definition, can't we straight away use the sequential criterion to prove this?
$f:Xto R^2$
$f(x)= (f_1(x),f_2(x))$ is continuous on X
$iff$
$forall langle x_n rangle in X $ such that $langle x_n rangle to ain X$
we have $f(x_n)= (f_1(x_n),f_2(x_n)) to (f_1(a),f_2(a)) ; ; forall ain X$
$iff$
$f_1(x_n) to f1(a)$ and $f_2(x_n) to f_2(a)$
$iff$
Both $f_1$ and $f_2 $ are continuous on X.
general-topology continuity metric-spaces
general-topology continuity metric-spaces
edited Nov 25 '18 at 5:21
So Lo
asked Nov 25 '18 at 5:20
So LoSo Lo
62719
62719
$begingroup$
You proof is surely valid.
$endgroup$
– Kavi Rama Murthy
Nov 25 '18 at 5:21
$begingroup$
you can also consider messing about with inclusions and projections, again to avoid writing much anything down
$endgroup$
– qbert
Nov 25 '18 at 5:24
add a comment |
$begingroup$
You proof is surely valid.
$endgroup$
– Kavi Rama Murthy
Nov 25 '18 at 5:21
$begingroup$
you can also consider messing about with inclusions and projections, again to avoid writing much anything down
$endgroup$
– qbert
Nov 25 '18 at 5:24
$begingroup$
You proof is surely valid.
$endgroup$
– Kavi Rama Murthy
Nov 25 '18 at 5:21
$begingroup$
You proof is surely valid.
$endgroup$
– Kavi Rama Murthy
Nov 25 '18 at 5:21
$begingroup$
you can also consider messing about with inclusions and projections, again to avoid writing much anything down
$endgroup$
– qbert
Nov 25 '18 at 5:24
$begingroup$
you can also consider messing about with inclusions and projections, again to avoid writing much anything down
$endgroup$
– qbert
Nov 25 '18 at 5:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Why is it true that $$(f_1(x_n), f_2(x_n)) rightarrow (f_1(a), f_2(a)) $$ if and only if $$f_1(x_n) rightarrow f_a(a) text{ and } f_s(x_n) rightarrow f_2(a)$$
If you've already proved this somewhere previously then good, you're re-using that proof to make your current proof simpler. But if you haven't already proved it then you've skipped over some important details, which are roughly equivalent to grinding out the $epsilon$-$delta$ proof.
$endgroup$
$begingroup$
I will use the result that convergence in $R^n$ is component wise
$endgroup$
– So Lo
Nov 25 '18 at 6:06
add a comment |
$begingroup$
Such a proof is indeed valid as soon as you know (and have proved) the following facts:
$f$ is continuous iff $f$ is sequentially continuous (this holds in metric spaces, and is commonly used in analysis).- A sequence $(a_n, b_n)_n $ in $mathbb{R}^2$ is convergent to $(a,b)$ iff $a_n to a$ and $b_n to b$.
Then still you could slightly rewrite it to prove the full equivalence (prove both directions separately):
Suppose $f_1$ and $f_2$ are continuous, to see that $f$ is (sequentially) continuous, let $(x_n)_n$ be a sequence in $X$ converging to $x in X$.
We then know by continuity of $f_1$ that $f(x_n) to f_1(x)$ and $f_2(x) to f_2(x)$, and so by 2. we know that $f(x_n) = (f_1(x_n), f_2(x_n)) to (f_1(x), f_2(x)) = f(x)$, as required. (This implication used the backward implication of 1. and the forward one of 2.)
Now suppose $f$ is continuous. We want to show $f_1$ and $f_2$ are continuous. let $x_n to x$ in $X$ again and we must show that $f_1(x_n) to f_1(x)$ and also $f_2(x_n) to f_2(x)$. Observe that $$(f_1(x_n), f_2(x_n)) = f(x_n) to f(x)= (f_1(x), f_2(x))$$ by sequential continuity of $f$ and so by the other direction of 2. we have that $f_1(x_n) to f_1(x)$ and also $f_2(x_n) to f_2(x)$, as required. (This implication used the forward implication of 1. and the backward one of 2.)
But of course the result itself is entirely independent of either metrics or sequences or even the finiteness of the product (just $2$ spaces here). In general topology this is just a consequence of the universal property of the product in the category of topological spaces.
$endgroup$
$begingroup$
I've proved the result in previous chapter that convergence in $R^n$ is component wise.
$endgroup$
– So Lo
Nov 25 '18 at 7:26
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%2f3012467%2ffx-f-1x-f-2x-is-continuous-on-x-iff-f-1-and-f-2-are-continuou%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
$begingroup$
Why is it true that $$(f_1(x_n), f_2(x_n)) rightarrow (f_1(a), f_2(a)) $$ if and only if $$f_1(x_n) rightarrow f_a(a) text{ and } f_s(x_n) rightarrow f_2(a)$$
If you've already proved this somewhere previously then good, you're re-using that proof to make your current proof simpler. But if you haven't already proved it then you've skipped over some important details, which are roughly equivalent to grinding out the $epsilon$-$delta$ proof.
$endgroup$
$begingroup$
I will use the result that convergence in $R^n$ is component wise
$endgroup$
– So Lo
Nov 25 '18 at 6:06
add a comment |
$begingroup$
Why is it true that $$(f_1(x_n), f_2(x_n)) rightarrow (f_1(a), f_2(a)) $$ if and only if $$f_1(x_n) rightarrow f_a(a) text{ and } f_s(x_n) rightarrow f_2(a)$$
If you've already proved this somewhere previously then good, you're re-using that proof to make your current proof simpler. But if you haven't already proved it then you've skipped over some important details, which are roughly equivalent to grinding out the $epsilon$-$delta$ proof.
$endgroup$
$begingroup$
I will use the result that convergence in $R^n$ is component wise
$endgroup$
– So Lo
Nov 25 '18 at 6:06
add a comment |
$begingroup$
Why is it true that $$(f_1(x_n), f_2(x_n)) rightarrow (f_1(a), f_2(a)) $$ if and only if $$f_1(x_n) rightarrow f_a(a) text{ and } f_s(x_n) rightarrow f_2(a)$$
If you've already proved this somewhere previously then good, you're re-using that proof to make your current proof simpler. But if you haven't already proved it then you've skipped over some important details, which are roughly equivalent to grinding out the $epsilon$-$delta$ proof.
$endgroup$
Why is it true that $$(f_1(x_n), f_2(x_n)) rightarrow (f_1(a), f_2(a)) $$ if and only if $$f_1(x_n) rightarrow f_a(a) text{ and } f_s(x_n) rightarrow f_2(a)$$
If you've already proved this somewhere previously then good, you're re-using that proof to make your current proof simpler. But if you haven't already proved it then you've skipped over some important details, which are roughly equivalent to grinding out the $epsilon$-$delta$ proof.
answered Nov 25 '18 at 5:39
JonathanZJonathanZ
2,119613
2,119613
$begingroup$
I will use the result that convergence in $R^n$ is component wise
$endgroup$
– So Lo
Nov 25 '18 at 6:06
add a comment |
$begingroup$
I will use the result that convergence in $R^n$ is component wise
$endgroup$
– So Lo
Nov 25 '18 at 6:06
$begingroup$
I will use the result that convergence in $R^n$ is component wise
$endgroup$
– So Lo
Nov 25 '18 at 6:06
$begingroup$
I will use the result that convergence in $R^n$ is component wise
$endgroup$
– So Lo
Nov 25 '18 at 6:06
add a comment |
$begingroup$
Such a proof is indeed valid as soon as you know (and have proved) the following facts:
$f$ is continuous iff $f$ is sequentially continuous (this holds in metric spaces, and is commonly used in analysis).- A sequence $(a_n, b_n)_n $ in $mathbb{R}^2$ is convergent to $(a,b)$ iff $a_n to a$ and $b_n to b$.
Then still you could slightly rewrite it to prove the full equivalence (prove both directions separately):
Suppose $f_1$ and $f_2$ are continuous, to see that $f$ is (sequentially) continuous, let $(x_n)_n$ be a sequence in $X$ converging to $x in X$.
We then know by continuity of $f_1$ that $f(x_n) to f_1(x)$ and $f_2(x) to f_2(x)$, and so by 2. we know that $f(x_n) = (f_1(x_n), f_2(x_n)) to (f_1(x), f_2(x)) = f(x)$, as required. (This implication used the backward implication of 1. and the forward one of 2.)
Now suppose $f$ is continuous. We want to show $f_1$ and $f_2$ are continuous. let $x_n to x$ in $X$ again and we must show that $f_1(x_n) to f_1(x)$ and also $f_2(x_n) to f_2(x)$. Observe that $$(f_1(x_n), f_2(x_n)) = f(x_n) to f(x)= (f_1(x), f_2(x))$$ by sequential continuity of $f$ and so by the other direction of 2. we have that $f_1(x_n) to f_1(x)$ and also $f_2(x_n) to f_2(x)$, as required. (This implication used the forward implication of 1. and the backward one of 2.)
But of course the result itself is entirely independent of either metrics or sequences or even the finiteness of the product (just $2$ spaces here). In general topology this is just a consequence of the universal property of the product in the category of topological spaces.
$endgroup$
$begingroup$
I've proved the result in previous chapter that convergence in $R^n$ is component wise.
$endgroup$
– So Lo
Nov 25 '18 at 7:26
add a comment |
$begingroup$
Such a proof is indeed valid as soon as you know (and have proved) the following facts:
$f$ is continuous iff $f$ is sequentially continuous (this holds in metric spaces, and is commonly used in analysis).- A sequence $(a_n, b_n)_n $ in $mathbb{R}^2$ is convergent to $(a,b)$ iff $a_n to a$ and $b_n to b$.
Then still you could slightly rewrite it to prove the full equivalence (prove both directions separately):
Suppose $f_1$ and $f_2$ are continuous, to see that $f$ is (sequentially) continuous, let $(x_n)_n$ be a sequence in $X$ converging to $x in X$.
We then know by continuity of $f_1$ that $f(x_n) to f_1(x)$ and $f_2(x) to f_2(x)$, and so by 2. we know that $f(x_n) = (f_1(x_n), f_2(x_n)) to (f_1(x), f_2(x)) = f(x)$, as required. (This implication used the backward implication of 1. and the forward one of 2.)
Now suppose $f$ is continuous. We want to show $f_1$ and $f_2$ are continuous. let $x_n to x$ in $X$ again and we must show that $f_1(x_n) to f_1(x)$ and also $f_2(x_n) to f_2(x)$. Observe that $$(f_1(x_n), f_2(x_n)) = f(x_n) to f(x)= (f_1(x), f_2(x))$$ by sequential continuity of $f$ and so by the other direction of 2. we have that $f_1(x_n) to f_1(x)$ and also $f_2(x_n) to f_2(x)$, as required. (This implication used the forward implication of 1. and the backward one of 2.)
But of course the result itself is entirely independent of either metrics or sequences or even the finiteness of the product (just $2$ spaces here). In general topology this is just a consequence of the universal property of the product in the category of topological spaces.
$endgroup$
$begingroup$
I've proved the result in previous chapter that convergence in $R^n$ is component wise.
$endgroup$
– So Lo
Nov 25 '18 at 7:26
add a comment |
$begingroup$
Such a proof is indeed valid as soon as you know (and have proved) the following facts:
$f$ is continuous iff $f$ is sequentially continuous (this holds in metric spaces, and is commonly used in analysis).- A sequence $(a_n, b_n)_n $ in $mathbb{R}^2$ is convergent to $(a,b)$ iff $a_n to a$ and $b_n to b$.
Then still you could slightly rewrite it to prove the full equivalence (prove both directions separately):
Suppose $f_1$ and $f_2$ are continuous, to see that $f$ is (sequentially) continuous, let $(x_n)_n$ be a sequence in $X$ converging to $x in X$.
We then know by continuity of $f_1$ that $f(x_n) to f_1(x)$ and $f_2(x) to f_2(x)$, and so by 2. we know that $f(x_n) = (f_1(x_n), f_2(x_n)) to (f_1(x), f_2(x)) = f(x)$, as required. (This implication used the backward implication of 1. and the forward one of 2.)
Now suppose $f$ is continuous. We want to show $f_1$ and $f_2$ are continuous. let $x_n to x$ in $X$ again and we must show that $f_1(x_n) to f_1(x)$ and also $f_2(x_n) to f_2(x)$. Observe that $$(f_1(x_n), f_2(x_n)) = f(x_n) to f(x)= (f_1(x), f_2(x))$$ by sequential continuity of $f$ and so by the other direction of 2. we have that $f_1(x_n) to f_1(x)$ and also $f_2(x_n) to f_2(x)$, as required. (This implication used the forward implication of 1. and the backward one of 2.)
But of course the result itself is entirely independent of either metrics or sequences or even the finiteness of the product (just $2$ spaces here). In general topology this is just a consequence of the universal property of the product in the category of topological spaces.
$endgroup$
Such a proof is indeed valid as soon as you know (and have proved) the following facts:
$f$ is continuous iff $f$ is sequentially continuous (this holds in metric spaces, and is commonly used in analysis).- A sequence $(a_n, b_n)_n $ in $mathbb{R}^2$ is convergent to $(a,b)$ iff $a_n to a$ and $b_n to b$.
Then still you could slightly rewrite it to prove the full equivalence (prove both directions separately):
Suppose $f_1$ and $f_2$ are continuous, to see that $f$ is (sequentially) continuous, let $(x_n)_n$ be a sequence in $X$ converging to $x in X$.
We then know by continuity of $f_1$ that $f(x_n) to f_1(x)$ and $f_2(x) to f_2(x)$, and so by 2. we know that $f(x_n) = (f_1(x_n), f_2(x_n)) to (f_1(x), f_2(x)) = f(x)$, as required. (This implication used the backward implication of 1. and the forward one of 2.)
Now suppose $f$ is continuous. We want to show $f_1$ and $f_2$ are continuous. let $x_n to x$ in $X$ again and we must show that $f_1(x_n) to f_1(x)$ and also $f_2(x_n) to f_2(x)$. Observe that $$(f_1(x_n), f_2(x_n)) = f(x_n) to f(x)= (f_1(x), f_2(x))$$ by sequential continuity of $f$ and so by the other direction of 2. we have that $f_1(x_n) to f_1(x)$ and also $f_2(x_n) to f_2(x)$, as required. (This implication used the forward implication of 1. and the backward one of 2.)
But of course the result itself is entirely independent of either metrics or sequences or even the finiteness of the product (just $2$ spaces here). In general topology this is just a consequence of the universal property of the product in the category of topological spaces.
answered Nov 25 '18 at 6:48
Henno BrandsmaHenno Brandsma
106k347114
106k347114
$begingroup$
I've proved the result in previous chapter that convergence in $R^n$ is component wise.
$endgroup$
– So Lo
Nov 25 '18 at 7:26
add a comment |
$begingroup$
I've proved the result in previous chapter that convergence in $R^n$ is component wise.
$endgroup$
– So Lo
Nov 25 '18 at 7:26
$begingroup$
I've proved the result in previous chapter that convergence in $R^n$ is component wise.
$endgroup$
– So Lo
Nov 25 '18 at 7:26
$begingroup$
I've proved the result in previous chapter that convergence in $R^n$ is component wise.
$endgroup$
– So Lo
Nov 25 '18 at 7:26
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%2f3012467%2ffx-f-1x-f-2x-is-continuous-on-x-iff-f-1-and-f-2-are-continuou%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
$begingroup$
You proof is surely valid.
$endgroup$
– Kavi Rama Murthy
Nov 25 '18 at 5:21
$begingroup$
you can also consider messing about with inclusions and projections, again to avoid writing much anything down
$endgroup$
– qbert
Nov 25 '18 at 5:24