Find a continuous surjection from $mathbb{R}^2$ to $Y = {(x, y) : 0 < x leq 1, y = sin (1/x)}$












0












$begingroup$


I'm looking for $f:mathbb{R}^2rightarrow Y$, where $f$ is a continuous surjection in the context of the following problem:



-Show the subspace $Y = {(x, y) : 0 < x leq 1, y = sin (1/x)}$ of $mathbb{R}^2$ is connected.



My solution strategy:



I want to apply the theorem that says: Let $f:(X, tau)rightarrow (Y,tau_1)$ be a surjective, continuous function from one topological space to another. Then, if $X$ is connected, so is $Y$.



Question: Will the following function work:



$$f(x,y)= left{ begin{array}{lcc}
(x, sin(1/x)), & 0 < x leq 1\
\ emptyset , &mbox{ otherwise,i.e., not defined}
\
end{array}right.$$



Clearly, this is a surjection since each element in $Y$ has an inverse image in $mathbb{R}^2$.



But I'm unsure about how to rigorously establish that $f$ is continuous by means of the general topological definition of a continuous function as a map from open sets to open sets.



Intuitively, I can envisage a correspondence between open rectangles in $mathbb{R}^2$ whose intersection with the infinite vertical strip $(0,1]times(-infty, infty)$ is mapped vertically to a connected portion of the curve $y=sin(1/x)$, and the inverse map where connected portions of the curve are mapped to the infinite vertical strip that covers its x-values. But this seems vague and hand-wavy.



Am I on the right track?



[P.S.: Apologies for one more question relating to the Topologist's Sine Curve, but I believe this one is different in its focus on continuity rather than connectedness:



Citation: S. Morris "Topology without Tears", 5.2.6 (i)]






general-topology continuity proof-writing







share|cite|improve this question











$endgroup$












  • $begingroup$
    What you defined is not a function $f:mathbb{R}^2to Y$. $f$ is only defined on a subset of $mathbb{R}$.
    $endgroup$
    – Eclipse Sun
    Dec 6 '18 at 22:23










  • $begingroup$
    Ok, then let's just use the subspace (0,1]×(−∞,∞) as the domain and skip the piecewise portion.
    $endgroup$
    – Cassius12
    Dec 6 '18 at 22:26










  • $begingroup$
    That's fine, and what you need to do next is to show $f$ is continuous. Do you know the fact that restriction of codomain preserves continuity? That is, if $f$ is a continuous map into $mathbb{R}^2$, and the image of $f$ is contained in $Y$, then $f$ is continuous as a map into $Y$.
    $endgroup$
    – Eclipse Sun
    Dec 6 '18 at 22:36










  • $begingroup$
    If you really wanted an answer to the title question, you could use something like $(x, y) mapsto (frac{1}{1+x^2}, sin(1+x^2))$ or $(x, y) mapsto e^{-x^2}, sin(e^{x^2}))$.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 23:30
















0












$begingroup$


I'm looking for $f:mathbb{R}^2rightarrow Y$, where $f$ is a continuous surjection in the context of the following problem:



-Show the subspace $Y = {(x, y) : 0 < x leq 1, y = sin (1/x)}$ of $mathbb{R}^2$ is connected.



My solution strategy:



I want to apply the theorem that says: Let $f:(X, tau)rightarrow (Y,tau_1)$ be a surjective, continuous function from one topological space to another. Then, if $X$ is connected, so is $Y$.



Question: Will the following function work:



$$f(x,y)= left{ begin{array}{lcc}
(x, sin(1/x)), & 0 < x leq 1\
\ emptyset , &mbox{ otherwise,i.e., not defined}
\
end{array}right.$$



Clearly, this is a surjection since each element in $Y$ has an inverse image in $mathbb{R}^2$.



But I'm unsure about how to rigorously establish that $f$ is continuous by means of the general topological definition of a continuous function as a map from open sets to open sets.



Intuitively, I can envisage a correspondence between open rectangles in $mathbb{R}^2$ whose intersection with the infinite vertical strip $(0,1]times(-infty, infty)$ is mapped vertically to a connected portion of the curve $y=sin(1/x)$, and the inverse map where connected portions of the curve are mapped to the infinite vertical strip that covers its x-values. But this seems vague and hand-wavy.



Am I on the right track?



[P.S.: Apologies for one more question relating to the Topologist's Sine Curve, but I believe this one is different in its focus on continuity rather than connectedness:



Citation: S. Morris "Topology without Tears", 5.2.6 (i)]






general-topology continuity proof-writing







share|cite|improve this question











$endgroup$












  • $begingroup$
    What you defined is not a function $f:mathbb{R}^2to Y$. $f$ is only defined on a subset of $mathbb{R}$.
    $endgroup$
    – Eclipse Sun
    Dec 6 '18 at 22:23










  • $begingroup$
    Ok, then let's just use the subspace (0,1]×(−∞,∞) as the domain and skip the piecewise portion.
    $endgroup$
    – Cassius12
    Dec 6 '18 at 22:26










  • $begingroup$
    That's fine, and what you need to do next is to show $f$ is continuous. Do you know the fact that restriction of codomain preserves continuity? That is, if $f$ is a continuous map into $mathbb{R}^2$, and the image of $f$ is contained in $Y$, then $f$ is continuous as a map into $Y$.
    $endgroup$
    – Eclipse Sun
    Dec 6 '18 at 22:36










  • $begingroup$
    If you really wanted an answer to the title question, you could use something like $(x, y) mapsto (frac{1}{1+x^2}, sin(1+x^2))$ or $(x, y) mapsto e^{-x^2}, sin(e^{x^2}))$.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 23:30














0












0








0





$begingroup$


I'm looking for $f:mathbb{R}^2rightarrow Y$, where $f$ is a continuous surjection in the context of the following problem:



-Show the subspace $Y = {(x, y) : 0 < x leq 1, y = sin (1/x)}$ of $mathbb{R}^2$ is connected.



My solution strategy:



I want to apply the theorem that says: Let $f:(X, tau)rightarrow (Y,tau_1)$ be a surjective, continuous function from one topological space to another. Then, if $X$ is connected, so is $Y$.



Question: Will the following function work:



$$f(x,y)= left{ begin{array}{lcc}
(x, sin(1/x)), & 0 < x leq 1\
\ emptyset , &mbox{ otherwise,i.e., not defined}
\
end{array}right.$$



Clearly, this is a surjection since each element in $Y$ has an inverse image in $mathbb{R}^2$.



But I'm unsure about how to rigorously establish that $f$ is continuous by means of the general topological definition of a continuous function as a map from open sets to open sets.



Intuitively, I can envisage a correspondence between open rectangles in $mathbb{R}^2$ whose intersection with the infinite vertical strip $(0,1]times(-infty, infty)$ is mapped vertically to a connected portion of the curve $y=sin(1/x)$, and the inverse map where connected portions of the curve are mapped to the infinite vertical strip that covers its x-values. But this seems vague and hand-wavy.



Am I on the right track?



[P.S.: Apologies for one more question relating to the Topologist's Sine Curve, but I believe this one is different in its focus on continuity rather than connectedness:



Citation: S. Morris "Topology without Tears", 5.2.6 (i)]






general-topology continuity proof-writing







share|cite|improve this question











$endgroup$




I'm looking for $f:mathbb{R}^2rightarrow Y$, where $f$ is a continuous surjection in the context of the following problem:



-Show the subspace $Y = {(x, y) : 0 < x leq 1, y = sin (1/x)}$ of $mathbb{R}^2$ is connected.



My solution strategy:



I want to apply the theorem that says: Let $f:(X, tau)rightarrow (Y,tau_1)$ be a surjective, continuous function from one topological space to another. Then, if $X$ is connected, so is $Y$.



Question: Will the following function work:



$$f(x,y)= left{ begin{array}{lcc}
(x, sin(1/x)), & 0 < x leq 1\
\ emptyset , &mbox{ otherwise,i.e., not defined}
\
end{array}right.$$



Clearly, this is a surjection since each element in $Y$ has an inverse image in $mathbb{R}^2$.



But I'm unsure about how to rigorously establish that $f$ is continuous by means of the general topological definition of a continuous function as a map from open sets to open sets.



Intuitively, I can envisage a correspondence between open rectangles in $mathbb{R}^2$ whose intersection with the infinite vertical strip $(0,1]times(-infty, infty)$ is mapped vertically to a connected portion of the curve $y=sin(1/x)$, and the inverse map where connected portions of the curve are mapped to the infinite vertical strip that covers its x-values. But this seems vague and hand-wavy.



Am I on the right track?



[P.S.: Apologies for one more question relating to the Topologist's Sine Curve, but I believe this one is different in its focus on continuity rather than connectedness:



Citation: S. Morris "Topology without Tears", 5.2.6 (i)]






general-topology continuity proof-writing




general-topology continuity proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 22:32







Cassius12

















asked Dec 6 '18 at 22:20









Cassius12Cassius12

11611




11611












  • $begingroup$
    What you defined is not a function $f:mathbb{R}^2to Y$. $f$ is only defined on a subset of $mathbb{R}$.
    $endgroup$
    – Eclipse Sun
    Dec 6 '18 at 22:23










  • $begingroup$
    Ok, then let's just use the subspace (0,1]×(−∞,∞) as the domain and skip the piecewise portion.
    $endgroup$
    – Cassius12
    Dec 6 '18 at 22:26










  • $begingroup$
    That's fine, and what you need to do next is to show $f$ is continuous. Do you know the fact that restriction of codomain preserves continuity? That is, if $f$ is a continuous map into $mathbb{R}^2$, and the image of $f$ is contained in $Y$, then $f$ is continuous as a map into $Y$.
    $endgroup$
    – Eclipse Sun
    Dec 6 '18 at 22:36










  • $begingroup$
    If you really wanted an answer to the title question, you could use something like $(x, y) mapsto (frac{1}{1+x^2}, sin(1+x^2))$ or $(x, y) mapsto e^{-x^2}, sin(e^{x^2}))$.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 23:30


















  • $begingroup$
    What you defined is not a function $f:mathbb{R}^2to Y$. $f$ is only defined on a subset of $mathbb{R}$.
    $endgroup$
    – Eclipse Sun
    Dec 6 '18 at 22:23










  • $begingroup$
    Ok, then let's just use the subspace (0,1]×(−∞,∞) as the domain and skip the piecewise portion.
    $endgroup$
    – Cassius12
    Dec 6 '18 at 22:26










  • $begingroup$
    That's fine, and what you need to do next is to show $f$ is continuous. Do you know the fact that restriction of codomain preserves continuity? That is, if $f$ is a continuous map into $mathbb{R}^2$, and the image of $f$ is contained in $Y$, then $f$ is continuous as a map into $Y$.
    $endgroup$
    – Eclipse Sun
    Dec 6 '18 at 22:36










  • $begingroup$
    If you really wanted an answer to the title question, you could use something like $(x, y) mapsto (frac{1}{1+x^2}, sin(1+x^2))$ or $(x, y) mapsto e^{-x^2}, sin(e^{x^2}))$.
    $endgroup$
    – Daniel Schepler
    Dec 6 '18 at 23:30
















$begingroup$
What you defined is not a function $f:mathbb{R}^2to Y$. $f$ is only defined on a subset of $mathbb{R}$.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 22:23




$begingroup$
What you defined is not a function $f:mathbb{R}^2to Y$. $f$ is only defined on a subset of $mathbb{R}$.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 22:23












$begingroup$
Ok, then let's just use the subspace (0,1]×(−∞,∞) as the domain and skip the piecewise portion.
$endgroup$
– Cassius12
Dec 6 '18 at 22:26




$begingroup$
Ok, then let's just use the subspace (0,1]×(−∞,∞) as the domain and skip the piecewise portion.
$endgroup$
– Cassius12
Dec 6 '18 at 22:26












$begingroup$
That's fine, and what you need to do next is to show $f$ is continuous. Do you know the fact that restriction of codomain preserves continuity? That is, if $f$ is a continuous map into $mathbb{R}^2$, and the image of $f$ is contained in $Y$, then $f$ is continuous as a map into $Y$.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 22:36




$begingroup$
That's fine, and what you need to do next is to show $f$ is continuous. Do you know the fact that restriction of codomain preserves continuity? That is, if $f$ is a continuous map into $mathbb{R}^2$, and the image of $f$ is contained in $Y$, then $f$ is continuous as a map into $Y$.
$endgroup$
– Eclipse Sun
Dec 6 '18 at 22:36












$begingroup$
If you really wanted an answer to the title question, you could use something like $(x, y) mapsto (frac{1}{1+x^2}, sin(1+x^2))$ or $(x, y) mapsto e^{-x^2}, sin(e^{x^2}))$.
$endgroup$
– Daniel Schepler
Dec 6 '18 at 23:30




$begingroup$
If you really wanted an answer to the title question, you could use something like $(x, y) mapsto (frac{1}{1+x^2}, sin(1+x^2))$ or $(x, y) mapsto e^{-x^2}, sin(e^{x^2}))$.
$endgroup$
– Daniel Schepler
Dec 6 '18 at 23:30










1 Answer
1






active

oldest

votes


















1












$begingroup$

I think it's easier to just take the domain to be $(0,1]$. Hopefully, the surjection (bijection in fact) to $Y$ is self-evident. Crucially, in your problem you don't have to worry about $x=0$, which is the real thorn for the Topologist's sine curve.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, wait, I see. I think I'm making things more difficult than they have to be. Are you saying, since the x-axis interval (0,1] x {0} is a subset of R^2, then that interval itself can be made into a subspace which I can then use as my domain to bijectively map the points (x,0) to (x, sin(1/x))?
    $endgroup$
    – Cassius12
    Dec 6 '18 at 22:45












  • $begingroup$
    Yes, basically. But I don't think it's that essential that (0,1] is a subspace of $mathbb{R}^2$. Image of a connected space under a continuous map is connected in general.
    $endgroup$
    – maridia
    Dec 6 '18 at 22:46












  • $begingroup$
    Ok, I agree. Now that I re-read the theorem I want to apply, I was incorrectly thinking that since Y was a subset of R^2, then X also had to be. But that's a false assumption -- (0,1] with the normal Euclidean topology is fine for X with no further details.
    $endgroup$
    – Cassius12
    Dec 6 '18 at 22:52











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

I think it's easier to just take the domain to be $(0,1]$. Hopefully, the surjection (bijection in fact) to $Y$ is self-evident. Crucially, in your problem you don't have to worry about $x=0$, which is the real thorn for the Topologist's sine curve.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, wait, I see. I think I'm making things more difficult than they have to be. Are you saying, since the x-axis interval (0,1] x {0} is a subset of R^2, then that interval itself can be made into a subspace which I can then use as my domain to bijectively map the points (x,0) to (x, sin(1/x))?
    $endgroup$
    – Cassius12
    Dec 6 '18 at 22:45












  • $begingroup$
    Yes, basically. But I don't think it's that essential that (0,1] is a subspace of $mathbb{R}^2$. Image of a connected space under a continuous map is connected in general.
    $endgroup$
    – maridia
    Dec 6 '18 at 22:46












  • $begingroup$
    Ok, I agree. Now that I re-read the theorem I want to apply, I was incorrectly thinking that since Y was a subset of R^2, then X also had to be. But that's a false assumption -- (0,1] with the normal Euclidean topology is fine for X with no further details.
    $endgroup$
    – Cassius12
    Dec 6 '18 at 22:52
















1












$begingroup$

I think it's easier to just take the domain to be $(0,1]$. Hopefully, the surjection (bijection in fact) to $Y$ is self-evident. Crucially, in your problem you don't have to worry about $x=0$, which is the real thorn for the Topologist's sine curve.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh, wait, I see. I think I'm making things more difficult than they have to be. Are you saying, since the x-axis interval (0,1] x {0} is a subset of R^2, then that interval itself can be made into a subspace which I can then use as my domain to bijectively map the points (x,0) to (x, sin(1/x))?
    $endgroup$
    – Cassius12
    Dec 6 '18 at 22:45












  • $begingroup$
    Yes, basically. But I don't think it's that essential that (0,1] is a subspace of $mathbb{R}^2$. Image of a connected space under a continuous map is connected in general.
    $endgroup$
    – maridia
    Dec 6 '18 at 22:46












  • $begingroup$
    Ok, I agree. Now that I re-read the theorem I want to apply, I was incorrectly thinking that since Y was a subset of R^2, then X also had to be. But that's a false assumption -- (0,1] with the normal Euclidean topology is fine for X with no further details.
    $endgroup$
    – Cassius12
    Dec 6 '18 at 22:52














1












1








1





$begingroup$

I think it's easier to just take the domain to be $(0,1]$. Hopefully, the surjection (bijection in fact) to $Y$ is self-evident. Crucially, in your problem you don't have to worry about $x=0$, which is the real thorn for the Topologist's sine curve.






share|cite|improve this answer









$endgroup$



I think it's easier to just take the domain to be $(0,1]$. Hopefully, the surjection (bijection in fact) to $Y$ is self-evident. Crucially, in your problem you don't have to worry about $x=0$, which is the real thorn for the Topologist's sine curve.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 22:35









maridiamaridia

1,065113




1,065113












  • $begingroup$
    Oh, wait, I see. I think I'm making things more difficult than they have to be. Are you saying, since the x-axis interval (0,1] x {0} is a subset of R^2, then that interval itself can be made into a subspace which I can then use as my domain to bijectively map the points (x,0) to (x, sin(1/x))?
    $endgroup$
    – Cassius12
    Dec 6 '18 at 22:45












  • $begingroup$
    Yes, basically. But I don't think it's that essential that (0,1] is a subspace of $mathbb{R}^2$. Image of a connected space under a continuous map is connected in general.
    $endgroup$
    – maridia
    Dec 6 '18 at 22:46












  • $begingroup$
    Ok, I agree. Now that I re-read the theorem I want to apply, I was incorrectly thinking that since Y was a subset of R^2, then X also had to be. But that's a false assumption -- (0,1] with the normal Euclidean topology is fine for X with no further details.
    $endgroup$
    – Cassius12
    Dec 6 '18 at 22:52


















  • $begingroup$
    Oh, wait, I see. I think I'm making things more difficult than they have to be. Are you saying, since the x-axis interval (0,1] x {0} is a subset of R^2, then that interval itself can be made into a subspace which I can then use as my domain to bijectively map the points (x,0) to (x, sin(1/x))?
    $endgroup$
    – Cassius12
    Dec 6 '18 at 22:45












  • $begingroup$
    Yes, basically. But I don't think it's that essential that (0,1] is a subspace of $mathbb{R}^2$. Image of a connected space under a continuous map is connected in general.
    $endgroup$
    – maridia
    Dec 6 '18 at 22:46












  • $begingroup$
    Ok, I agree. Now that I re-read the theorem I want to apply, I was incorrectly thinking that since Y was a subset of R^2, then X also had to be. But that's a false assumption -- (0,1] with the normal Euclidean topology is fine for X with no further details.
    $endgroup$
    – Cassius12
    Dec 6 '18 at 22:52
















$begingroup$
Oh, wait, I see. I think I'm making things more difficult than they have to be. Are you saying, since the x-axis interval (0,1] x {0} is a subset of R^2, then that interval itself can be made into a subspace which I can then use as my domain to bijectively map the points (x,0) to (x, sin(1/x))?
$endgroup$
– Cassius12
Dec 6 '18 at 22:45






$begingroup$
Oh, wait, I see. I think I'm making things more difficult than they have to be. Are you saying, since the x-axis interval (0,1] x {0} is a subset of R^2, then that interval itself can be made into a subspace which I can then use as my domain to bijectively map the points (x,0) to (x, sin(1/x))?
$endgroup$
– Cassius12
Dec 6 '18 at 22:45














$begingroup$
Yes, basically. But I don't think it's that essential that (0,1] is a subspace of $mathbb{R}^2$. Image of a connected space under a continuous map is connected in general.
$endgroup$
– maridia
Dec 6 '18 at 22:46






$begingroup$
Yes, basically. But I don't think it's that essential that (0,1] is a subspace of $mathbb{R}^2$. Image of a connected space under a continuous map is connected in general.
$endgroup$
– maridia
Dec 6 '18 at 22:46














$begingroup$
Ok, I agree. Now that I re-read the theorem I want to apply, I was incorrectly thinking that since Y was a subset of R^2, then X also had to be. But that's a false assumption -- (0,1] with the normal Euclidean topology is fine for X with no further details.
$endgroup$
– Cassius12
Dec 6 '18 at 22:52




$begingroup$
Ok, I agree. Now that I re-read the theorem I want to apply, I was incorrectly thinking that since Y was a subset of R^2, then X also had to be. But that's a false assumption -- (0,1] with the normal Euclidean topology is fine for X with no further details.
$endgroup$
– Cassius12
Dec 6 '18 at 22:52


















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Berounka

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Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
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Sphinx de Gizeh

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Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
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Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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