$f(x)= (f_1(x),f_2(x))$ is continuous on X $iff $ $f_1$ and $f_2$ are continuous












1












$begingroup$



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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















1












$begingroup$



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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












1








1





$begingroup$



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.










share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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




















  • $begingroup$
    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












2 Answers
2






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I will use the result that convergence in $R^n$ is component wise
    $endgroup$
    – So Lo
    Nov 25 '18 at 6:06



















0












$begingroup$

Such a proof is indeed valid as soon as you know (and have proved) the following facts:





  1. $f$ is continuous iff $f$ is sequentially continuous (this holds in metric spaces, and is commonly used in analysis).

  2. 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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I will use the result that convergence in $R^n$ is component wise
    $endgroup$
    – So Lo
    Nov 25 '18 at 6:06
















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I will use the result that convergence in $R^n$ is component wise
    $endgroup$
    – So Lo
    Nov 25 '18 at 6:06














0












0








0





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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











0












$begingroup$

Such a proof is indeed valid as soon as you know (and have proved) the following facts:





  1. $f$ is continuous iff $f$ is sequentially continuous (this holds in metric spaces, and is commonly used in analysis).

  2. 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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

Such a proof is indeed valid as soon as you know (and have proved) the following facts:





  1. $f$ is continuous iff $f$ is sequentially continuous (this holds in metric spaces, and is commonly used in analysis).

  2. 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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

Such a proof is indeed valid as soon as you know (and have proved) the following facts:





  1. $f$ is continuous iff $f$ is sequentially continuous (this holds in metric spaces, and is commonly used in analysis).

  2. 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.






share|cite|improve this answer









$endgroup$



Such a proof is indeed valid as soon as you know (and have proved) the following facts:





  1. $f$ is continuous iff $f$ is sequentially continuous (this holds in metric spaces, and is commonly used in analysis).

  2. 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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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