Homeomorphisms on $mathbb{R}$
Consider the following topologies on $mathbb{R}$:
$tau_1 = {mathbb{R}, emptyset, (-n, n) forall nin mathbb{Z}^+}$
$tau_2 = {mathbb{R}, emptyset, [-n, n] forall nin mathbb{Z}^+}$
$tau_3 = {mathbb{R}, emptyset, (-r, r) forall rin mathbb{R}^+}$
$tau_4 = {mathbb{R}, emptyset, (-r, r)$ and $[-r, r] forall rin mathbb{R}^+}$
Problem 1: Show $tau_1 notcong tau_3$
Problem 2: Is $tau_1 cong tau_2$?
Problem 3: Is $tau_3 cong tau_4$?
Note for beginners: Do not assume facts about $mathbb{R}$ based on the standard Euclidean topology. These are different topologies, and so they may behave differently on otherwise familiar sets-- for example, $(a,b)$ and $[a,b]$ are non-homeomorphic interval types in the Euclidean topology, but you can't use that here. Work from the definition of Homeomorphism.
Citation: S. Morris, "Topology without Tears", problem 4.3.7 i, ii, iii
general-topology
asked Nov 24 at 18:28
Cassius12
10611
add a comment |
Consider the following topologies on $mathbb{R}$:
$tau_1 = {mathbb{R}, emptyset, (-n, n) forall nin mathbb{Z}^+}$
$tau_2 = {mathbb{R}, emptyset, [-n, n] forall nin mathbb{Z}^+}$
$tau_3 = {mathbb{R}, emptyset, (-r, r) forall rin mathbb{R}^+}$
$tau_4 = {mathbb{R}, emptyset, (-r, r)$ and $[-r, r] forall rin mathbb{R}^+}$
Problem 1: Show $tau_1 notcong tau_3$
Problem 2: Is $tau_1 cong tau_2$?
Problem 3: Is $tau_3 cong tau_4$?
Note for beginners: Do not assume facts about $mathbb{R}$ based on the standard Euclidean topology. These are different topologies, and so they may behave differently on otherwise familiar sets-- for example, $(a,b)$ and $[a,b]$ are non-homeomorphic interval types in the Euclidean topology, but you can't use that here. Work from the definition of Homeomorphism.
Citation: S. Morris, "Topology without Tears", problem 4.3.7 i, ii, iii
general-topology
asked Nov 24 at 18:28
Cassius12
10611
By the way, the material that you put in your deleted answer should have been part of the question itself: it always makes for a better question if you include your own attempted solutions.
– Lee Mosher
Nov 26 at 16:54
ok, I put that in my note for beginners.
– Cassius12
Nov 27 at 2:32
I think you misunderstood my previous comment. What I meant was this: When you are writing a question, and when you have developed partial answers to your own question, but you are unsure about those answers, put those partial answers into your question. It makes for a better question. So, for example, your partial answers to 1), to 2), and to 3) should have been part of the original question.
– Lee Mosher
Nov 27 at 13:12
add a comment |
Consider the following topologies on $mathbb{R}$:
$tau_1 = {mathbb{R}, emptyset, (-n, n) forall nin mathbb{Z}^+}$
$tau_2 = {mathbb{R}, emptyset, [-n, n] forall nin mathbb{Z}^+}$
$tau_3 = {mathbb{R}, emptyset, (-r, r) forall rin mathbb{R}^+}$
$tau_4 = {mathbb{R}, emptyset, (-r, r)$ and $[-r, r] forall rin mathbb{R}^+}$
Problem 1: Show $tau_1 notcong tau_3$
Problem 2: Is $tau_1 cong tau_2$?
Problem 3: Is $tau_3 cong tau_4$?
Note for beginners: Do not assume facts about $mathbb{R}$ based on the standard Euclidean topology. These are different topologies, and so they may behave differently on otherwise familiar sets-- for example, $(a,b)$ and $[a,b]$ are non-homeomorphic interval types in the Euclidean topology, but you can't use that here. Work from the definition of Homeomorphism.
Citation: S. Morris, "Topology without Tears", problem 4.3.7 i, ii, iii
general-topology
asked Nov 24 at 18:28
Cassius12
10611
Consider the following topologies on $mathbb{R}$:
$tau_1 = {mathbb{R}, emptyset, (-n, n) forall nin mathbb{Z}^+}$
$tau_2 = {mathbb{R}, emptyset, [-n, n] forall nin mathbb{Z}^+}$
$tau_3 = {mathbb{R}, emptyset, (-r, r) forall rin mathbb{R}^+}$
$tau_4 = {mathbb{R}, emptyset, (-r, r)$ and $[-r, r] forall rin mathbb{R}^+}$
Problem 1: Show $tau_1 notcong tau_3$
Problem 2: Is $tau_1 cong tau_2$?
Problem 3: Is $tau_3 cong tau_4$?
Note for beginners: Do not assume facts about $mathbb{R}$ based on the standard Euclidean topology. These are different topologies, and so they may behave differently on otherwise familiar sets-- for example, $(a,b)$ and $[a,b]$ are non-homeomorphic interval types in the Euclidean topology, but you can't use that here. Work from the definition of Homeomorphism.
Citation: S. Morris, "Topology without Tears", problem 4.3.7 i, ii, iii
general-topology
general-topology
asked Nov 24 at 18:28
Cassius12
10611
asked Nov 24 at 18:28
Cassius12
10611
edited Nov 27 at 2:31
asked Nov 24 at 18:28
Cassius12
10611
asked Nov 24 at 18:28
Cassius12
10611
asked Nov 24 at 18:28
Cassius12
10611
10611
By the way, the material that you put in your deleted answer should have been part of the question itself: it always makes for a better question if you include your own attempted solutions.
– Lee Mosher
Nov 26 at 16:54
ok, I put that in my note for beginners.
– Cassius12
Nov 27 at 2:32
I think you misunderstood my previous comment. What I meant was this: When you are writing a question, and when you have developed partial answers to your own question, but you are unsure about those answers, put those partial answers into your question. It makes for a better question. So, for example, your partial answers to 1), to 2), and to 3) should have been part of the original question.
– Lee Mosher
Nov 27 at 13:12
add a comment |
By the way, the material that you put in your deleted answer should have been part of the question itself: it always makes for a better question if you include your own attempted solutions.
– Lee Mosher
Nov 26 at 16:54
ok, I put that in my note for beginners.
– Cassius12
Nov 27 at 2:32
I think you misunderstood my previous comment. What I meant was this: When you are writing a question, and when you have developed partial answers to your own question, but you are unsure about those answers, put those partial answers into your question. It makes for a better question. So, for example, your partial answers to 1), to 2), and to 3) should have been part of the original question.
– Lee Mosher
Nov 27 at 13:12
By the way, the material that you put in your deleted answer should have been part of the question itself: it always makes for a better question if you include your own attempted solutions.
– Lee Mosher
Nov 26 at 16:54
By the way, the material that you put in your deleted answer should have been part of the question itself: it always makes for a better question if you include your own attempted solutions.
– Lee Mosher
Nov 26 at 16:54
ok, I put that in my note for beginners.
– Cassius12
Nov 27 at 2:32
ok, I put that in my note for beginners.
– Cassius12
Nov 27 at 2:32
I think you misunderstood my previous comment. What I meant was this: When you are writing a question, and when you have developed partial answers to your own question, but you are unsure about those answers, put those partial answers into your question. It makes for a better question. So, for example, your partial answers to 1), to 2), and to 3) should have been part of the original question.
– Lee Mosher
Nov 27 at 13:12
I think you misunderstood my previous comment. What I meant was this: When you are writing a question, and when you have developed partial answers to your own question, but you are unsure about those answers, put those partial answers into your question. It makes for a better question. So, for example, your partial answers to 1), to 2), and to 3) should have been part of the original question.
– Lee Mosher
Nov 27 at 13:12
add a comment |
1 Answer
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$|tau_1| neq |tau_3|$ while a homeomorphism induces a bijection of topologies.
Let $f_1$ be a bijection between $(-1,1)$ and $[-1,1]$ (standard set theory).
We can extend this to a bijection $f_2$ between $(-2,2)$ and $[-2,2]$ (both difference sets are also equinumerous).
So continuing this way, we get a sequence of extensions that are bijections between $(-n,n)$ and $[-n,n]$. The union of these functions is then a bijection of the reals that for each $n$ obeys $f[(-n,n)] = [-n,n]$ (and so also $f^{-1}[-n,n]] = (-n,n)$) and so $f$ is both open and continuous. So $tau_1 simeq tau_2$.
In $tau_4$ the open sets have the property:
$$exists U,V in tau_4: U subsetneq V land |V setminus U| =2tag{1}$$
While $tau_3$ does not have the above property as it obeys the property
$$forall U,V in tau_3: U subsetneq V implies |V setminus U| text{ is infinite}tag{2}$$
So $tau_3 not simeq tau_4$:
If $f: (mathbb{R}, tau_4) to (mathbb{R}, tau_3)$ would be a homeomorphism, then for $U,V$ as in $(1)$ we'd have $f[U] subsetneq f[V]$, both would be open and $f[V setminus U] = f[V] setminus f[U]$ would contradict $(2)$.
answered Nov 24 at 18:53
Henno Brandsma
105k346113
add a comment |
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$|tau_1| neq |tau_3|$ while a homeomorphism induces a bijection of topologies.
Let $f_1$ be a bijection between $(-1,1)$ and $[-1,1]$ (standard set theory).
We can extend this to a bijection $f_2$ between $(-2,2)$ and $[-2,2]$ (both difference sets are also equinumerous).
So continuing this way, we get a sequence of extensions that are bijections between $(-n,n)$ and $[-n,n]$. The union of these functions is then a bijection of the reals that for each $n$ obeys $f[(-n,n)] = [-n,n]$ (and so also $f^{-1}[-n,n]] = (-n,n)$) and so $f$ is both open and continuous. So $tau_1 simeq tau_2$.
In $tau_4$ the open sets have the property:
$$exists U,V in tau_4: U subsetneq V land |V setminus U| =2tag{1}$$
While $tau_3$ does not have the above property as it obeys the property
$$forall U,V in tau_3: U subsetneq V implies |V setminus U| text{ is infinite}tag{2}$$
So $tau_3 not simeq tau_4$:
If $f: (mathbb{R}, tau_4) to (mathbb{R}, tau_3)$ would be a homeomorphism, then for $U,V$ as in $(1)$ we'd have $f[U] subsetneq f[V]$, both would be open and $f[V setminus U] = f[V] setminus f[U]$ would contradict $(2)$.
answered Nov 24 at 18:53
Henno Brandsma
105k346113
add a comment |
$|tau_1| neq |tau_3|$ while a homeomorphism induces a bijection of topologies.
Let $f_1$ be a bijection between $(-1,1)$ and $[-1,1]$ (standard set theory).
We can extend this to a bijection $f_2$ between $(-2,2)$ and $[-2,2]$ (both difference sets are also equinumerous).
So continuing this way, we get a sequence of extensions that are bijections between $(-n,n)$ and $[-n,n]$. The union of these functions is then a bijection of the reals that for each $n$ obeys $f[(-n,n)] = [-n,n]$ (and so also $f^{-1}[-n,n]] = (-n,n)$) and so $f$ is both open and continuous. So $tau_1 simeq tau_2$.
In $tau_4$ the open sets have the property:
$$exists U,V in tau_4: U subsetneq V land |V setminus U| =2tag{1}$$
While $tau_3$ does not have the above property as it obeys the property
$$forall U,V in tau_3: U subsetneq V implies |V setminus U| text{ is infinite}tag{2}$$
So $tau_3 not simeq tau_4$:
If $f: (mathbb{R}, tau_4) to (mathbb{R}, tau_3)$ would be a homeomorphism, then for $U,V$ as in $(1)$ we'd have $f[U] subsetneq f[V]$, both would be open and $f[V setminus U] = f[V] setminus f[U]$ would contradict $(2)$.
answered Nov 24 at 18:53
Henno Brandsma
105k346113
add a comment |
$|tau_1| neq |tau_3|$ while a homeomorphism induces a bijection of topologies.
Let $f_1$ be a bijection between $(-1,1)$ and $[-1,1]$ (standard set theory).
We can extend this to a bijection $f_2$ between $(-2,2)$ and $[-2,2]$ (both difference sets are also equinumerous).
So continuing this way, we get a sequence of extensions that are bijections between $(-n,n)$ and $[-n,n]$. The union of these functions is then a bijection of the reals that for each $n$ obeys $f[(-n,n)] = [-n,n]$ (and so also $f^{-1}[-n,n]] = (-n,n)$) and so $f$ is both open and continuous. So $tau_1 simeq tau_2$.
In $tau_4$ the open sets have the property:
$$exists U,V in tau_4: U subsetneq V land |V setminus U| =2tag{1}$$
While $tau_3$ does not have the above property as it obeys the property
$$forall U,V in tau_3: U subsetneq V implies |V setminus U| text{ is infinite}tag{2}$$
So $tau_3 not simeq tau_4$:
If $f: (mathbb{R}, tau_4) to (mathbb{R}, tau_3)$ would be a homeomorphism, then for $U,V$ as in $(1)$ we'd have $f[U] subsetneq f[V]$, both would be open and $f[V setminus U] = f[V] setminus f[U]$ would contradict $(2)$.
answered Nov 24 at 18:53
Henno Brandsma
105k346113
$|tau_1| neq |tau_3|$ while a homeomorphism induces a bijection of topologies.
Let $f_1$ be a bijection between $(-1,1)$ and $[-1,1]$ (standard set theory).
We can extend this to a bijection $f_2$ between $(-2,2)$ and $[-2,2]$ (both difference sets are also equinumerous).
So continuing this way, we get a sequence of extensions that are bijections between $(-n,n)$ and $[-n,n]$. The union of these functions is then a bijection of the reals that for each $n$ obeys $f[(-n,n)] = [-n,n]$ (and so also $f^{-1}[-n,n]] = (-n,n)$) and so $f$ is both open and continuous. So $tau_1 simeq tau_2$.
In $tau_4$ the open sets have the property:
$$exists U,V in tau_4: U subsetneq V land |V setminus U| =2tag{1}$$
While $tau_3$ does not have the above property as it obeys the property
$$forall U,V in tau_3: U subsetneq V implies |V setminus U| text{ is infinite}tag{2}$$
So $tau_3 not simeq tau_4$:
If $f: (mathbb{R}, tau_4) to (mathbb{R}, tau_3)$ would be a homeomorphism, then for $U,V$ as in $(1)$ we'd have $f[U] subsetneq f[V]$, both would be open and $f[V setminus U] = f[V] setminus f[U]$ would contradict $(2)$.
answered Nov 24 at 18:53
Henno Brandsma
105k346113
edited Nov 24 at 18:59
answered Nov 24 at 18:53
Henno Brandsma
105k346113
answered Nov 24 at 18:53
Henno Brandsma
105k346113
answered Nov 24 at 18:53
Henno Brandsma
105k346113
105k346113
add a comment |
add a comment |
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By the way, the material that you put in your deleted answer should have been part of the question itself: it always makes for a better question if you include your own attempted solutions.
– Lee Mosher
Nov 26 at 16:54
ok, I put that in my note for beginners.
– Cassius12
Nov 27 at 2:32
I think you misunderstood my previous comment. What I meant was this: When you are writing a question, and when you have developed partial answers to your own question, but you are unsure about those answers, put those partial answers into your question. It makes for a better question. So, for example, your partial answers to 1), to 2), and to 3) should have been part of the original question.
– Lee Mosher
Nov 27 at 13:12