Find a continuous surjection from $mathbb{R}^2$ to $Y = {(x, y) : 0 < x leq 1, y = sin (1/x)}$
$begingroup$
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
asked Dec 6 '18 at 22:20
Cassius12Cassius12
13111
$endgroup$
add a comment |
$begingroup$
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
asked Dec 6 '18 at 22:20
Cassius12Cassius12
13111
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 6 '18 at 22:20
Cassius12Cassius12
13111
$endgroup$
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
asked Dec 6 '18 at 22:20
Cassius12Cassius12
13111
asked Dec 6 '18 at 22:20
Cassius12Cassius12
13111
edited Dec 6 '18 at 22:32
Cassius12
asked Dec 6 '18 at 22:20
Cassius12Cassius12
13111
asked Dec 6 '18 at 22:20
Cassius12Cassius12
13111
asked Dec 6 '18 at 22:20
Cassius12Cassius12
13111
13111
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 6 '18 at 22:35
zoidbergzoidberg
1,070113
$endgroup$
$begingroup$
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.
$endgroup$
– zoidberg
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
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
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.
answered Dec 6 '18 at 22:35
zoidbergzoidberg
1,070113
$endgroup$
$begingroup$
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.
$endgroup$
– zoidberg
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
add a comment |
$begingroup$
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.
answered Dec 6 '18 at 22:35
zoidbergzoidberg
1,070113
$endgroup$
$begingroup$
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.
$endgroup$
– zoidberg
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
add a comment |
$begingroup$
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.
answered Dec 6 '18 at 22:35
zoidbergzoidberg
1,070113
$endgroup$
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.
answered Dec 6 '18 at 22:35
zoidbergzoidberg
1,070113
answered Dec 6 '18 at 22:35
zoidbergzoidberg
1,070113
answered Dec 6 '18 at 22:35
zoidbergzoidberg
1,070113
answered Dec 6 '18 at 22:35
zoidbergzoidberg
1,070113
1,070113
$begingroup$
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.
$endgroup$
– zoidberg
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
add a comment |
$begingroup$
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.
$endgroup$
– zoidberg
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.
$endgroup$
– zoidberg
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.
$endgroup$
– zoidberg
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
add a comment |
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$begingroup$
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 $(x, y) mapsto e^{-x^2}, sin(e^{x^2}))$.
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– Daniel Schepler
Dec 6 '18 at 23:30