$X$ is homeomorphic to $Xtimes X$ (TIFR GS $2014$)
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Question is :
Suppose $X$ is a topological space of infinite cardinality which is homeomorphic to $Xtimes X$. Then which of the following is true:
- $X$ is not connected.
- $X$ is not compact
- $X$ is not homeomorphic to a subset of $mathbb{R}$
- None of the above.
I guess first two options are false.
We do have possibility that product of two connected spaces is connected.
So, $Xtimes X$ is connected if $X$ is connected. So I guess there is no problem.
We do have possibility that product of two compact spaces is compact.
So, $Xtimes X$ is compact if $X$ is compact. So I guess there is no problem.
I understand that this is not the proof to exclude first two options but I guess the chance is more for them to be false.
So, only thing I have problem with is third option.
I could do nothing for that third option..
I would be thankful if some one can help me out to clear this.
Thank you :)
general-topology examples-counterexamples product-space
edited Nov 19 '16 at 16:56
Martin Sleziak
44.8k10119273
$endgroup$
add a comment |
$begingroup$
Question is :
Suppose $X$ is a topological space of infinite cardinality which is homeomorphic to $Xtimes X$. Then which of the following is true:
- $X$ is not connected.
- $X$ is not compact
- $X$ is not homeomorphic to a subset of $mathbb{R}$
- None of the above.
I guess first two options are false.
We do have possibility that product of two connected spaces is connected.
So, $Xtimes X$ is connected if $X$ is connected. So I guess there is no problem.
We do have possibility that product of two compact spaces is compact.
So, $Xtimes X$ is compact if $X$ is compact. So I guess there is no problem.
I understand that this is not the proof to exclude first two options but I guess the chance is more for them to be false.
So, only thing I have problem with is third option.
I could do nothing for that third option..
I would be thankful if some one can help me out to clear this.
Thank you :)
general-topology examples-counterexamples product-space
edited Nov 19 '16 at 16:56
Martin Sleziak
44.8k10119273
$endgroup$
$begingroup$
$mathbb N$ is a simple counterexample for the third one.
$endgroup$
– Dejan Govc
Dec 10 '13 at 10:24
$begingroup$
$[0,1]^infty{}{}$
$endgroup$
– user98602
Dec 12 '15 at 7:06
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@MikeMiller Drat, beat me to it! (As a good further exercise, it's easy to show that you can't get a simultaneous counterexample to $A$ and $C$.)
$endgroup$
– Noah Schweber
Dec 12 '15 at 7:08
1
$begingroup$
@NoahSchweber: I had the advantage of not using words :)
$endgroup$
– user98602
Dec 12 '15 at 7:08
$begingroup$
Why $mathbb{Q}$ in the discrete topology? The usual subspace topology, works, but it's not quite trivial to see that $mathbb{Q} times mathbb{Q} approx mathbb{Q}$, but $mathbb{Z}$ in the discrete = subspace topology is a better example (admittedly homeomorphic to the rationals in the discrete topology, but the latter does not directly embed into the reals). The irrationals or the Cantor set also work, etc.
$endgroup$
– Henno Brandsma
Dec 12 '15 at 8:37
add a comment |
$begingroup$
Question is :
Suppose $X$ is a topological space of infinite cardinality which is homeomorphic to $Xtimes X$. Then which of the following is true:
- $X$ is not connected.
- $X$ is not compact
- $X$ is not homeomorphic to a subset of $mathbb{R}$
- None of the above.
I guess first two options are false.
We do have possibility that product of two connected spaces is connected.
So, $Xtimes X$ is connected if $X$ is connected. So I guess there is no problem.
We do have possibility that product of two compact spaces is compact.
So, $Xtimes X$ is compact if $X$ is compact. So I guess there is no problem.
I understand that this is not the proof to exclude first two options but I guess the chance is more for them to be false.
So, only thing I have problem with is third option.
I could do nothing for that third option..
I would be thankful if some one can help me out to clear this.
Thank you :)
general-topology examples-counterexamples product-space
edited Nov 19 '16 at 16:56
Martin Sleziak
44.8k10119273
$endgroup$
Question is :
Suppose $X$ is a topological space of infinite cardinality which is homeomorphic to $Xtimes X$. Then which of the following is true:
- $X$ is not connected.
- $X$ is not compact
- $X$ is not homeomorphic to a subset of $mathbb{R}$
- None of the above.
I guess first two options are false.
We do have possibility that product of two connected spaces is connected.
So, $Xtimes X$ is connected if $X$ is connected. So I guess there is no problem.
We do have possibility that product of two compact spaces is compact.
So, $Xtimes X$ is compact if $X$ is compact. So I guess there is no problem.
I understand that this is not the proof to exclude first two options but I guess the chance is more for them to be false.
So, only thing I have problem with is third option.
I could do nothing for that third option..
I would be thankful if some one can help me out to clear this.
Thank you :)
general-topology examples-counterexamples product-space
general-topology examples-counterexamples product-space
edited Nov 19 '16 at 16:56
Martin Sleziak
44.8k10119273
edited Nov 19 '16 at 16:56
Martin Sleziak
44.8k10119273
edited Nov 19 '16 at 16:56
Martin Sleziak
44.8k10119273
edited Nov 19 '16 at 16:56
Martin Sleziak
44.8k10119273
edited Nov 19 '16 at 16:56
Martin Sleziak
44.8k10119273
44.8k10119273
asked Dec 10 '13 at 8:56
user87543
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$mathbb N$ is a simple counterexample for the third one.
$endgroup$
– Dejan Govc
Dec 10 '13 at 10:24
$begingroup$
$[0,1]^infty{}{}$
$endgroup$
– user98602
Dec 12 '15 at 7:06
$begingroup$
@MikeMiller Drat, beat me to it! (As a good further exercise, it's easy to show that you can't get a simultaneous counterexample to $A$ and $C$.)
$endgroup$
– Noah Schweber
Dec 12 '15 at 7:08
1
$begingroup$
@NoahSchweber: I had the advantage of not using words :)
$endgroup$
– user98602
Dec 12 '15 at 7:08
$begingroup$
Why $mathbb{Q}$ in the discrete topology? The usual subspace topology, works, but it's not quite trivial to see that $mathbb{Q} times mathbb{Q} approx mathbb{Q}$, but $mathbb{Z}$ in the discrete = subspace topology is a better example (admittedly homeomorphic to the rationals in the discrete topology, but the latter does not directly embed into the reals). The irrationals or the Cantor set also work, etc.
$endgroup$
– Henno Brandsma
Dec 12 '15 at 8:37
add a comment |
$begingroup$
$mathbb N$ is a simple counterexample for the third one.
$endgroup$
– Dejan Govc
Dec 10 '13 at 10:24
$begingroup$
$[0,1]^infty{}{}$
$endgroup$
– user98602
Dec 12 '15 at 7:06
$begingroup$
@MikeMiller Drat, beat me to it! (As a good further exercise, it's easy to show that you can't get a simultaneous counterexample to $A$ and $C$.)
$endgroup$
– Noah Schweber
Dec 12 '15 at 7:08
1
$begingroup$
@NoahSchweber: I had the advantage of not using words :)
$endgroup$
– user98602
Dec 12 '15 at 7:08
$begingroup$
Why $mathbb{Q}$ in the discrete topology? The usual subspace topology, works, but it's not quite trivial to see that $mathbb{Q} times mathbb{Q} approx mathbb{Q}$, but $mathbb{Z}$ in the discrete = subspace topology is a better example (admittedly homeomorphic to the rationals in the discrete topology, but the latter does not directly embed into the reals). The irrationals or the Cantor set also work, etc.
$endgroup$
– Henno Brandsma
Dec 12 '15 at 8:37
$begingroup$
$mathbb N$ is a simple counterexample for the third one.
$endgroup$
– Dejan Govc
Dec 10 '13 at 10:24
$begingroup$
$mathbb N$ is a simple counterexample for the third one.
$endgroup$
– Dejan Govc
Dec 10 '13 at 10:24
$begingroup$
$[0,1]^infty{}{}$
$endgroup$
– user98602
Dec 12 '15 at 7:06
$begingroup$
$[0,1]^infty{}{}$
$endgroup$
– user98602
Dec 12 '15 at 7:06
$begingroup$
@MikeMiller Drat, beat me to it! (As a good further exercise, it's easy to show that you can't get a simultaneous counterexample to $A$ and $C$.)
$endgroup$
– Noah Schweber
Dec 12 '15 at 7:08
$begingroup$
@MikeMiller Drat, beat me to it! (As a good further exercise, it's easy to show that you can't get a simultaneous counterexample to $A$ and $C$.)
$endgroup$
– Noah Schweber
Dec 12 '15 at 7:08
1
1
$begingroup$
@NoahSchweber: I had the advantage of not using words :)
$endgroup$
– user98602
Dec 12 '15 at 7:08
$begingroup$
@NoahSchweber: I had the advantage of not using words :)
$endgroup$
– user98602
Dec 12 '15 at 7:08
$begingroup$
Why $mathbb{Q}$ in the discrete topology? The usual subspace topology, works, but it's not quite trivial to see that $mathbb{Q} times mathbb{Q} approx mathbb{Q}$, but $mathbb{Z}$ in the discrete = subspace topology is a better example (admittedly homeomorphic to the rationals in the discrete topology, but the latter does not directly embed into the reals). The irrationals or the Cantor set also work, etc.
$endgroup$
– Henno Brandsma
Dec 12 '15 at 8:37
$begingroup$
Why $mathbb{Q}$ in the discrete topology? The usual subspace topology, works, but it's not quite trivial to see that $mathbb{Q} times mathbb{Q} approx mathbb{Q}$, but $mathbb{Z}$ in the discrete = subspace topology is a better example (admittedly homeomorphic to the rationals in the discrete topology, but the latter does not directly embed into the reals). The irrationals or the Cantor set also work, etc.
$endgroup$
– Henno Brandsma
Dec 12 '15 at 8:37
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The Cantor set is a counter-example to the second and third statement. Note that the Cantor set is homeomorphic to ${0,1}^{mathbb N}$, hence it is homeomorphic to the product with itself.
An infinite set with the smallest topology (exactly two open sets) is a counter-example to the first statement. Martini gives a better counter-example in a comment.
answered Dec 10 '13 at 9:03
Carsten SCarsten S
7,26811436
$endgroup$
2
$begingroup$
... of infinite cardinality. But $[0,1]^{mathbb N}$ works as counterexample to the first and second statement.
$endgroup$
– martini
Dec 10 '13 at 9:04
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@martini : about the third statement? :(
$endgroup$
– user87543
Dec 10 '13 at 9:06
$begingroup$
As said by martini, my space must be of infinite cardinality...
$endgroup$
– user87543
Dec 10 '13 at 9:06
$begingroup$
@martini, thanks I had overlooked that, I will correct this.
$endgroup$
– Carsten S
Dec 10 '13 at 9:07
1
$begingroup$
You are welcome. Cantor space is the first space I think of when I see $Xtimes Xapprox X$.
$endgroup$
– Carsten S
Dec 10 '13 at 9:17
|
show 5 more comments
$begingroup$
Fix your favorite compact connected space $X$ - then $X^omega$ is compact, connected, and homeomorphic to its own square.
answered Dec 12 '15 at 7:07
Noah SchweberNoah Schweber
126k10151290
$endgroup$
add a comment |
$begingroup$
Take the Cantor set. This gives a counterexample for B and C. For A, as Mike said in the comments, take the set $[0,1]^infty$.
answered Dec 12 '15 at 7:08
seekerseeker
1,189414
$endgroup$
add a comment |
$begingroup$
Let $Q = prod_{n=1}^infty I_n$ where $I_n = [0,1]$. This space is known as the Hilbert cube. It is compact and connected and satisfies $Q times Q approx Q$.
Note that you cannot find a single counterexample $X$ for a) and c). Assume $X$ would be such a space. Then $X$ must be homeomorphic to a connected subset of $mathbb{R}$, that is, homeomorphic to an interval $J$. But $J times J$ is not homeomorphic to $J$.
answered Dec 6 '18 at 4:28
jasminejasmine
1,845418
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The Cantor set is a counter-example to the second and third statement. Note that the Cantor set is homeomorphic to ${0,1}^{mathbb N}$, hence it is homeomorphic to the product with itself.
An infinite set with the smallest topology (exactly two open sets) is a counter-example to the first statement. Martini gives a better counter-example in a comment.
answered Dec 10 '13 at 9:03
Carsten SCarsten S
7,26811436
$endgroup$
2
$begingroup$
... of infinite cardinality. But $[0,1]^{mathbb N}$ works as counterexample to the first and second statement.
$endgroup$
– martini
Dec 10 '13 at 9:04
$begingroup$
@martini : about the third statement? :(
$endgroup$
– user87543
Dec 10 '13 at 9:06
$begingroup$
As said by martini, my space must be of infinite cardinality...
$endgroup$
– user87543
Dec 10 '13 at 9:06
$begingroup$
@martini, thanks I had overlooked that, I will correct this.
$endgroup$
– Carsten S
Dec 10 '13 at 9:07
1
$begingroup$
You are welcome. Cantor space is the first space I think of when I see $Xtimes Xapprox X$.
$endgroup$
– Carsten S
Dec 10 '13 at 9:17
|
show 5 more comments
$begingroup$
The Cantor set is a counter-example to the second and third statement. Note that the Cantor set is homeomorphic to ${0,1}^{mathbb N}$, hence it is homeomorphic to the product with itself.
An infinite set with the smallest topology (exactly two open sets) is a counter-example to the first statement. Martini gives a better counter-example in a comment.
answered Dec 10 '13 at 9:03
Carsten SCarsten S
7,26811436
$endgroup$
2
$begingroup$
... of infinite cardinality. But $[0,1]^{mathbb N}$ works as counterexample to the first and second statement.
$endgroup$
– martini
Dec 10 '13 at 9:04
$begingroup$
@martini : about the third statement? :(
$endgroup$
– user87543
Dec 10 '13 at 9:06
$begingroup$
As said by martini, my space must be of infinite cardinality...
$endgroup$
– user87543
Dec 10 '13 at 9:06
$begingroup$
@martini, thanks I had overlooked that, I will correct this.
$endgroup$
– Carsten S
Dec 10 '13 at 9:07
1
$begingroup$
You are welcome. Cantor space is the first space I think of when I see $Xtimes Xapprox X$.
$endgroup$
– Carsten S
Dec 10 '13 at 9:17
|
show 5 more comments
$begingroup$
The Cantor set is a counter-example to the second and third statement. Note that the Cantor set is homeomorphic to ${0,1}^{mathbb N}$, hence it is homeomorphic to the product with itself.
An infinite set with the smallest topology (exactly two open sets) is a counter-example to the first statement. Martini gives a better counter-example in a comment.
answered Dec 10 '13 at 9:03
Carsten SCarsten S
7,26811436
$endgroup$
The Cantor set is a counter-example to the second and third statement. Note that the Cantor set is homeomorphic to ${0,1}^{mathbb N}$, hence it is homeomorphic to the product with itself.
An infinite set with the smallest topology (exactly two open sets) is a counter-example to the first statement. Martini gives a better counter-example in a comment.
answered Dec 10 '13 at 9:03
Carsten SCarsten S
7,26811436
edited Dec 10 '13 at 9:10
answered Dec 10 '13 at 9:03
Carsten SCarsten S
7,26811436
answered Dec 10 '13 at 9:03
Carsten SCarsten S
7,26811436
answered Dec 10 '13 at 9:03
Carsten SCarsten S
7,26811436
7,26811436
2
$begingroup$
... of infinite cardinality. But $[0,1]^{mathbb N}$ works as counterexample to the first and second statement.
$endgroup$
– martini
Dec 10 '13 at 9:04
$begingroup$
@martini : about the third statement? :(
$endgroup$
– user87543
Dec 10 '13 at 9:06
$begingroup$
As said by martini, my space must be of infinite cardinality...
$endgroup$
– user87543
Dec 10 '13 at 9:06
$begingroup$
@martini, thanks I had overlooked that, I will correct this.
$endgroup$
– Carsten S
Dec 10 '13 at 9:07
1
$begingroup$
You are welcome. Cantor space is the first space I think of when I see $Xtimes Xapprox X$.
$endgroup$
– Carsten S
Dec 10 '13 at 9:17
|
show 5 more comments
2
$begingroup$
... of infinite cardinality. But $[0,1]^{mathbb N}$ works as counterexample to the first and second statement.
$endgroup$
– martini
Dec 10 '13 at 9:04
$begingroup$
@martini : about the third statement? :(
$endgroup$
– user87543
Dec 10 '13 at 9:06
$begingroup$
As said by martini, my space must be of infinite cardinality...
$endgroup$
– user87543
Dec 10 '13 at 9:06
$begingroup$
@martini, thanks I had overlooked that, I will correct this.
$endgroup$
– Carsten S
Dec 10 '13 at 9:07
1
$begingroup$
You are welcome. Cantor space is the first space I think of when I see $Xtimes Xapprox X$.
$endgroup$
– Carsten S
Dec 10 '13 at 9:17
2
2
$begingroup$
... of infinite cardinality. But $[0,1]^{mathbb N}$ works as counterexample to the first and second statement.
$endgroup$
– martini
Dec 10 '13 at 9:04
$begingroup$
... of infinite cardinality. But $[0,1]^{mathbb N}$ works as counterexample to the first and second statement.
$endgroup$
– martini
Dec 10 '13 at 9:04
$begingroup$
@martini : about the third statement? :(
$endgroup$
– user87543
Dec 10 '13 at 9:06
$begingroup$
@martini : about the third statement? :(
$endgroup$
– user87543
Dec 10 '13 at 9:06
$begingroup$
As said by martini, my space must be of infinite cardinality...
$endgroup$
– user87543
Dec 10 '13 at 9:06
$begingroup$
As said by martini, my space must be of infinite cardinality...
$endgroup$
– user87543
Dec 10 '13 at 9:06
$begingroup$
@martini, thanks I had overlooked that, I will correct this.
$endgroup$
– Carsten S
Dec 10 '13 at 9:07
$begingroup$
@martini, thanks I had overlooked that, I will correct this.
$endgroup$
– Carsten S
Dec 10 '13 at 9:07
1
1
$begingroup$
You are welcome. Cantor space is the first space I think of when I see $Xtimes Xapprox X$.
$endgroup$
– Carsten S
Dec 10 '13 at 9:17
$begingroup$
You are welcome. Cantor space is the first space I think of when I see $Xtimes Xapprox X$.
$endgroup$
– Carsten S
Dec 10 '13 at 9:17
|
show 5 more comments
$begingroup$
Fix your favorite compact connected space $X$ - then $X^omega$ is compact, connected, and homeomorphic to its own square.
answered Dec 12 '15 at 7:07
Noah SchweberNoah Schweber
126k10151290
$endgroup$
add a comment |
$begingroup$
Fix your favorite compact connected space $X$ - then $X^omega$ is compact, connected, and homeomorphic to its own square.
answered Dec 12 '15 at 7:07
Noah SchweberNoah Schweber
126k10151290
$endgroup$
add a comment |
$begingroup$
Fix your favorite compact connected space $X$ - then $X^omega$ is compact, connected, and homeomorphic to its own square.
answered Dec 12 '15 at 7:07
Noah SchweberNoah Schweber
126k10151290
$endgroup$
Fix your favorite compact connected space $X$ - then $X^omega$ is compact, connected, and homeomorphic to its own square.
answered Dec 12 '15 at 7:07
Noah SchweberNoah Schweber
126k10151290
answered Dec 12 '15 at 7:07
Noah SchweberNoah Schweber
126k10151290
answered Dec 12 '15 at 7:07
Noah SchweberNoah Schweber
126k10151290
answered Dec 12 '15 at 7:07
Noah SchweberNoah Schweber
126k10151290
126k10151290
add a comment |
add a comment |
$begingroup$
Take the Cantor set. This gives a counterexample for B and C. For A, as Mike said in the comments, take the set $[0,1]^infty$.
answered Dec 12 '15 at 7:08
seekerseeker
1,189414
$endgroup$
add a comment |
$begingroup$
Take the Cantor set. This gives a counterexample for B and C. For A, as Mike said in the comments, take the set $[0,1]^infty$.
answered Dec 12 '15 at 7:08
seekerseeker
1,189414
$endgroup$
add a comment |
$begingroup$
Take the Cantor set. This gives a counterexample for B and C. For A, as Mike said in the comments, take the set $[0,1]^infty$.
answered Dec 12 '15 at 7:08
seekerseeker
1,189414
$endgroup$
Take the Cantor set. This gives a counterexample for B and C. For A, as Mike said in the comments, take the set $[0,1]^infty$.
answered Dec 12 '15 at 7:08
seekerseeker
1,189414
answered Dec 12 '15 at 7:08
seekerseeker
1,189414
answered Dec 12 '15 at 7:08
seekerseeker
1,189414
answered Dec 12 '15 at 7:08
seekerseeker
1,189414
1,189414
add a comment |
add a comment |
$begingroup$
Let $Q = prod_{n=1}^infty I_n$ where $I_n = [0,1]$. This space is known as the Hilbert cube. It is compact and connected and satisfies $Q times Q approx Q$.
Note that you cannot find a single counterexample $X$ for a) and c). Assume $X$ would be such a space. Then $X$ must be homeomorphic to a connected subset of $mathbb{R}$, that is, homeomorphic to an interval $J$. But $J times J$ is not homeomorphic to $J$.
answered Dec 6 '18 at 4:28
jasminejasmine
1,845418
$endgroup$
add a comment |
$begingroup$
Let $Q = prod_{n=1}^infty I_n$ where $I_n = [0,1]$. This space is known as the Hilbert cube. It is compact and connected and satisfies $Q times Q approx Q$.
Note that you cannot find a single counterexample $X$ for a) and c). Assume $X$ would be such a space. Then $X$ must be homeomorphic to a connected subset of $mathbb{R}$, that is, homeomorphic to an interval $J$. But $J times J$ is not homeomorphic to $J$.
answered Dec 6 '18 at 4:28
jasminejasmine
1,845418
$endgroup$
add a comment |
$begingroup$
Let $Q = prod_{n=1}^infty I_n$ where $I_n = [0,1]$. This space is known as the Hilbert cube. It is compact and connected and satisfies $Q times Q approx Q$.
Note that you cannot find a single counterexample $X$ for a) and c). Assume $X$ would be such a space. Then $X$ must be homeomorphic to a connected subset of $mathbb{R}$, that is, homeomorphic to an interval $J$. But $J times J$ is not homeomorphic to $J$.
answered Dec 6 '18 at 4:28
jasminejasmine
1,845418
$endgroup$
Let $Q = prod_{n=1}^infty I_n$ where $I_n = [0,1]$. This space is known as the Hilbert cube. It is compact and connected and satisfies $Q times Q approx Q$.
Note that you cannot find a single counterexample $X$ for a) and c). Assume $X$ would be such a space. Then $X$ must be homeomorphic to a connected subset of $mathbb{R}$, that is, homeomorphic to an interval $J$. But $J times J$ is not homeomorphic to $J$.
answered Dec 6 '18 at 4:28
jasminejasmine
1,845418
answered Dec 6 '18 at 4:28
jasminejasmine
1,845418
answered Dec 6 '18 at 4:28
jasminejasmine
1,845418
answered Dec 6 '18 at 4:28
jasminejasmine
1,845418
1,845418
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$begingroup$
$mathbb N$ is a simple counterexample for the third one.
$endgroup$
– Dejan Govc
Dec 10 '13 at 10:24
$begingroup$
$[0,1]^infty{}{}$
$endgroup$
– user98602
Dec 12 '15 at 7:06
$begingroup$
@MikeMiller Drat, beat me to it! (As a good further exercise, it's easy to show that you can't get a simultaneous counterexample to $A$ and $C$.)
$endgroup$
– Noah Schweber
Dec 12 '15 at 7:08
1
$begingroup$
@NoahSchweber: I had the advantage of not using words :)
$endgroup$
– user98602
Dec 12 '15 at 7:08
$begingroup$
Why $mathbb{Q}$ in the discrete topology? The usual subspace topology, works, but it's not quite trivial to see that $mathbb{Q} times mathbb{Q} approx mathbb{Q}$, but $mathbb{Z}$ in the discrete = subspace topology is a better example (admittedly homeomorphic to the rationals in the discrete topology, but the latter does not directly embed into the reals). The irrationals or the Cantor set also work, etc.
$endgroup$
– Henno Brandsma
Dec 12 '15 at 8:37