$omega^omega$ correspondence with $mathbb R$
How does the natural continuous bijection between $omega^omega$ and $mathbb R$ look like? I.e. why elements of $omega^omega$ are called reals?
continuity real-numbers integers natural-numbers
asked Dec 26 '18 at 17:52
user122424
1,0881616
|
show 7 more comments
How does the natural continuous bijection between $omega^omega$ and $mathbb R$ look like? I.e. why elements of $omega^omega$ are called reals?
continuity real-numbers integers natural-numbers
asked Dec 26 '18 at 17:52
user122424
1,0881616
Who calls them that?
– Hagen von Eitzen
Dec 26 '18 at 17:57
In my experience one typically identifies $omega^omega$ with the irrational elements of $mathbb R$; and then we call them "reals" because they are equinumerous, and in particular "isomorphic up to a countable set".
– Mees de Vries
Dec 26 '18 at 18:01
In books on forcing they always use reals for $omega^omega$. Am I wrong?
– user122424
Dec 26 '18 at 18:01
@HagenvonEitzen Sadly, logicians do tend to refer to elements of Baire space, elements of Cantor space, and related objects as "reals." This is because all the relevant spaces are Borel-isomorphic, so in many (but not all) contexts they are essentially interchangeable.
– Noah Schweber
Dec 26 '18 at 18:01
@HagenvonEitzen, see e.g. Wikipedia, which states "This space is commonly used in descriptive set theory, to the extent that its elements are often called “reals.”"
– Mees de Vries
Dec 26 '18 at 18:02
|
show 7 more comments
How does the natural continuous bijection between $omega^omega$ and $mathbb R$ look like? I.e. why elements of $omega^omega$ are called reals?
continuity real-numbers integers natural-numbers
asked Dec 26 '18 at 17:52
user122424
1,0881616
How does the natural continuous bijection between $omega^omega$ and $mathbb R$ look like? I.e. why elements of $omega^omega$ are called reals?
continuity real-numbers integers natural-numbers
continuity real-numbers integers natural-numbers
asked Dec 26 '18 at 17:52
user122424
1,0881616
asked Dec 26 '18 at 17:52
user122424
1,0881616
asked Dec 26 '18 at 17:52
user122424
1,0881616
asked Dec 26 '18 at 17:52
user122424
1,0881616
asked Dec 26 '18 at 17:52
user122424
1,0881616
1,0881616
Who calls them that?
– Hagen von Eitzen
Dec 26 '18 at 17:57
In my experience one typically identifies $omega^omega$ with the irrational elements of $mathbb R$; and then we call them "reals" because they are equinumerous, and in particular "isomorphic up to a countable set".
– Mees de Vries
Dec 26 '18 at 18:01
In books on forcing they always use reals for $omega^omega$. Am I wrong?
– user122424
Dec 26 '18 at 18:01
@HagenvonEitzen Sadly, logicians do tend to refer to elements of Baire space, elements of Cantor space, and related objects as "reals." This is because all the relevant spaces are Borel-isomorphic, so in many (but not all) contexts they are essentially interchangeable.
– Noah Schweber
Dec 26 '18 at 18:01
@HagenvonEitzen, see e.g. Wikipedia, which states "This space is commonly used in descriptive set theory, to the extent that its elements are often called “reals.”"
– Mees de Vries
Dec 26 '18 at 18:02
|
show 7 more comments
Who calls them that?
– Hagen von Eitzen
Dec 26 '18 at 17:57
In my experience one typically identifies $omega^omega$ with the irrational elements of $mathbb R$; and then we call them "reals" because they are equinumerous, and in particular "isomorphic up to a countable set".
– Mees de Vries
Dec 26 '18 at 18:01
In books on forcing they always use reals for $omega^omega$. Am I wrong?
– user122424
Dec 26 '18 at 18:01
@HagenvonEitzen Sadly, logicians do tend to refer to elements of Baire space, elements of Cantor space, and related objects as "reals." This is because all the relevant spaces are Borel-isomorphic, so in many (but not all) contexts they are essentially interchangeable.
– Noah Schweber
Dec 26 '18 at 18:01
@HagenvonEitzen, see e.g. Wikipedia, which states "This space is commonly used in descriptive set theory, to the extent that its elements are often called “reals.”"
– Mees de Vries
Dec 26 '18 at 18:02
Who calls them that?
– Hagen von Eitzen
Dec 26 '18 at 17:57
Who calls them that?
– Hagen von Eitzen
Dec 26 '18 at 17:57
In my experience one typically identifies $omega^omega$ with the irrational elements of $mathbb R$; and then we call them "reals" because they are equinumerous, and in particular "isomorphic up to a countable set".
– Mees de Vries
Dec 26 '18 at 18:01
In my experience one typically identifies $omega^omega$ with the irrational elements of $mathbb R$; and then we call them "reals" because they are equinumerous, and in particular "isomorphic up to a countable set".
– Mees de Vries
Dec 26 '18 at 18:01
In books on forcing they always use reals for $omega^omega$. Am I wrong?
– user122424
Dec 26 '18 at 18:01
In books on forcing they always use reals for $omega^omega$. Am I wrong?
– user122424
Dec 26 '18 at 18:01
@HagenvonEitzen Sadly, logicians do tend to refer to elements of Baire space, elements of Cantor space, and related objects as "reals." This is because all the relevant spaces are Borel-isomorphic, so in many (but not all) contexts they are essentially interchangeable.
– Noah Schweber
Dec 26 '18 at 18:01
@HagenvonEitzen Sadly, logicians do tend to refer to elements of Baire space, elements of Cantor space, and related objects as "reals." This is because all the relevant spaces are Borel-isomorphic, so in many (but not all) contexts they are essentially interchangeable.
– Noah Schweber
Dec 26 '18 at 18:01
@HagenvonEitzen, see e.g. Wikipedia, which states "This space is commonly used in descriptive set theory, to the extent that its elements are often called “reals.”"
– Mees de Vries
Dec 26 '18 at 18:02
@HagenvonEitzen, see e.g. Wikipedia, which states "This space is commonly used in descriptive set theory, to the extent that its elements are often called “reals.”"
– Mees de Vries
Dec 26 '18 at 18:02
|
show 7 more comments
1 Answer
1
active
oldest
votes
Contra the fifth word of your question, there is no natural continuous bijection between $omega^omega$ and $mathbb{R}$. Indeed, there isn't even a continuous injection from $mathbb{R}$ to $omega^omega$ since the former is connected but the latter is totally disconnected.
However, there are still reasonably-low-complexity bijections between the two - namely, with respect to the natural topologies they are Borel isomorphic (that is, there is a Borel bijection between the two). We can see this by finding Borel injections in each direction, and then checking that the Cantor-Bernstein construction turns a pair of Borel injections into a Borel bijection.
Cooking up specific Borel injections each way is a good exercise. But here are a couple hints in each direction:
The map $bin$ sending a real in $(0,1)$ to its canonical (= not-eventually-all-$1$s) binary representation is an injection into $omega^omega$ (in fact, into $2^omega$). $(0,1)$ and $mathbb{R}$ are clearly homeomorphic; can you show that $bin$ is Borel? (Note that we know per the above that $bin$ can't be continuous - the discontinuity crops up at the dyadic reals, do you see why?)
In the other direction, the most commonly used map is via continued fraction expansions. However, it may be more intuitive to go a bit more combinatorial. First, we can go from $omega^omega$ to $2^omega$ by "counting $0$s:" given $finomega^omega$, we write a binary sequence such that the number of $0$s between the $n$th and $(n+1)$th $1$s is $f(n)$. As an example, $f=(1,4,3,0,2,...)$ goes to $$(1,0,1,0,0,0,0,1,0,0,0,1,1,0,0,1,...)$$ This is obviously an injection, and is easily checked to be Borel; now just compose with the usual continuous map from $2^omega$ to $mathbb{R}$.
Thinking along these lines, it's also easy to check that in fact $omega^omega$ is homeomorphic to the set of irrationals (again, each with the usual topology).
Introductory texts on descriptive set theory - e.g. Kechris and Moschovakis - go into this in more detail. In particular, this is a specific example of the more general fact that any two uncountable Polish spaces are Borel isomorphic.
All of this says that the difference between the reals and the reals (hehe) is fairly minor - we can conflate the two either via bijection of fairly low complexity, or by ignoring a small (= countable) set. This means that in many situations (e.g. forcing and descriptive set theory) it essentially doesn't matter which we use.
There are situations, of course, where the difference is meaningful - e.g. in computable structure theory (here/here) - but they are in practice rare enough that the abuse of terminology doesn't lead to trouble in practice.
answered Dec 26 '18 at 18:08
Noah Schweber
121k10148284
Can you explicitly construct the two Borel injections?
– user122424
Dec 26 '18 at 18:11
Yes. There are completely explicit examples of such injections.
– Andrés E. Caicedo
Dec 26 '18 at 18:14
@user122424 I've described them in a bit of detail - would you like me to flesh them out more?
– Noah Schweber
Dec 26 '18 at 18:14
@user122424 Also, note that the word "the" is unjustified - there's no unique feature of these! There are lots of reasonably-simple-to-describe Borel injections. Cooking more examples up is a good exercise, incidentally.
– Noah Schweber
Dec 26 '18 at 18:28
@user122424 Incidentally, given these questions I strongly recommend reading the first couple chapters of Kechris before continuing on with forcing - a good sense for how to "move between" essentially-equivalent spaces will help clarify things immensely.
– Noah Schweber
Dec 26 '18 at 18:38
|
show 3 more comments
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Contra the fifth word of your question, there is no natural continuous bijection between $omega^omega$ and $mathbb{R}$. Indeed, there isn't even a continuous injection from $mathbb{R}$ to $omega^omega$ since the former is connected but the latter is totally disconnected.
However, there are still reasonably-low-complexity bijections between the two - namely, with respect to the natural topologies they are Borel isomorphic (that is, there is a Borel bijection between the two). We can see this by finding Borel injections in each direction, and then checking that the Cantor-Bernstein construction turns a pair of Borel injections into a Borel bijection.
Cooking up specific Borel injections each way is a good exercise. But here are a couple hints in each direction:
The map $bin$ sending a real in $(0,1)$ to its canonical (= not-eventually-all-$1$s) binary representation is an injection into $omega^omega$ (in fact, into $2^omega$). $(0,1)$ and $mathbb{R}$ are clearly homeomorphic; can you show that $bin$ is Borel? (Note that we know per the above that $bin$ can't be continuous - the discontinuity crops up at the dyadic reals, do you see why?)
In the other direction, the most commonly used map is via continued fraction expansions. However, it may be more intuitive to go a bit more combinatorial. First, we can go from $omega^omega$ to $2^omega$ by "counting $0$s:" given $finomega^omega$, we write a binary sequence such that the number of $0$s between the $n$th and $(n+1)$th $1$s is $f(n)$. As an example, $f=(1,4,3,0,2,...)$ goes to $$(1,0,1,0,0,0,0,1,0,0,0,1,1,0,0,1,...)$$ This is obviously an injection, and is easily checked to be Borel; now just compose with the usual continuous map from $2^omega$ to $mathbb{R}$.
Thinking along these lines, it's also easy to check that in fact $omega^omega$ is homeomorphic to the set of irrationals (again, each with the usual topology).
Introductory texts on descriptive set theory - e.g. Kechris and Moschovakis - go into this in more detail. In particular, this is a specific example of the more general fact that any two uncountable Polish spaces are Borel isomorphic.
All of this says that the difference between the reals and the reals (hehe) is fairly minor - we can conflate the two either via bijection of fairly low complexity, or by ignoring a small (= countable) set. This means that in many situations (e.g. forcing and descriptive set theory) it essentially doesn't matter which we use.
There are situations, of course, where the difference is meaningful - e.g. in computable structure theory (here/here) - but they are in practice rare enough that the abuse of terminology doesn't lead to trouble in practice.
answered Dec 26 '18 at 18:08
Noah Schweber
121k10148284
Can you explicitly construct the two Borel injections?
– user122424
Dec 26 '18 at 18:11
Yes. There are completely explicit examples of such injections.
– Andrés E. Caicedo
Dec 26 '18 at 18:14
@user122424 I've described them in a bit of detail - would you like me to flesh them out more?
– Noah Schweber
Dec 26 '18 at 18:14
@user122424 Also, note that the word "the" is unjustified - there's no unique feature of these! There are lots of reasonably-simple-to-describe Borel injections. Cooking more examples up is a good exercise, incidentally.
– Noah Schweber
Dec 26 '18 at 18:28
@user122424 Incidentally, given these questions I strongly recommend reading the first couple chapters of Kechris before continuing on with forcing - a good sense for how to "move between" essentially-equivalent spaces will help clarify things immensely.
– Noah Schweber
Dec 26 '18 at 18:38
|
show 3 more comments
Contra the fifth word of your question, there is no natural continuous bijection between $omega^omega$ and $mathbb{R}$. Indeed, there isn't even a continuous injection from $mathbb{R}$ to $omega^omega$ since the former is connected but the latter is totally disconnected.
However, there are still reasonably-low-complexity bijections between the two - namely, with respect to the natural topologies they are Borel isomorphic (that is, there is a Borel bijection between the two). We can see this by finding Borel injections in each direction, and then checking that the Cantor-Bernstein construction turns a pair of Borel injections into a Borel bijection.
Cooking up specific Borel injections each way is a good exercise. But here are a couple hints in each direction:
The map $bin$ sending a real in $(0,1)$ to its canonical (= not-eventually-all-$1$s) binary representation is an injection into $omega^omega$ (in fact, into $2^omega$). $(0,1)$ and $mathbb{R}$ are clearly homeomorphic; can you show that $bin$ is Borel? (Note that we know per the above that $bin$ can't be continuous - the discontinuity crops up at the dyadic reals, do you see why?)
In the other direction, the most commonly used map is via continued fraction expansions. However, it may be more intuitive to go a bit more combinatorial. First, we can go from $omega^omega$ to $2^omega$ by "counting $0$s:" given $finomega^omega$, we write a binary sequence such that the number of $0$s between the $n$th and $(n+1)$th $1$s is $f(n)$. As an example, $f=(1,4,3,0,2,...)$ goes to $$(1,0,1,0,0,0,0,1,0,0,0,1,1,0,0,1,...)$$ This is obviously an injection, and is easily checked to be Borel; now just compose with the usual continuous map from $2^omega$ to $mathbb{R}$.
Thinking along these lines, it's also easy to check that in fact $omega^omega$ is homeomorphic to the set of irrationals (again, each with the usual topology).
Introductory texts on descriptive set theory - e.g. Kechris and Moschovakis - go into this in more detail. In particular, this is a specific example of the more general fact that any two uncountable Polish spaces are Borel isomorphic.
All of this says that the difference between the reals and the reals (hehe) is fairly minor - we can conflate the two either via bijection of fairly low complexity, or by ignoring a small (= countable) set. This means that in many situations (e.g. forcing and descriptive set theory) it essentially doesn't matter which we use.
There are situations, of course, where the difference is meaningful - e.g. in computable structure theory (here/here) - but they are in practice rare enough that the abuse of terminology doesn't lead to trouble in practice.
answered Dec 26 '18 at 18:08
Noah Schweber
121k10148284
Can you explicitly construct the two Borel injections?
– user122424
Dec 26 '18 at 18:11
Yes. There are completely explicit examples of such injections.
– Andrés E. Caicedo
Dec 26 '18 at 18:14
@user122424 I've described them in a bit of detail - would you like me to flesh them out more?
– Noah Schweber
Dec 26 '18 at 18:14
@user122424 Also, note that the word "the" is unjustified - there's no unique feature of these! There are lots of reasonably-simple-to-describe Borel injections. Cooking more examples up is a good exercise, incidentally.
– Noah Schweber
Dec 26 '18 at 18:28
@user122424 Incidentally, given these questions I strongly recommend reading the first couple chapters of Kechris before continuing on with forcing - a good sense for how to "move between" essentially-equivalent spaces will help clarify things immensely.
– Noah Schweber
Dec 26 '18 at 18:38
|
show 3 more comments
Contra the fifth word of your question, there is no natural continuous bijection between $omega^omega$ and $mathbb{R}$. Indeed, there isn't even a continuous injection from $mathbb{R}$ to $omega^omega$ since the former is connected but the latter is totally disconnected.
However, there are still reasonably-low-complexity bijections between the two - namely, with respect to the natural topologies they are Borel isomorphic (that is, there is a Borel bijection between the two). We can see this by finding Borel injections in each direction, and then checking that the Cantor-Bernstein construction turns a pair of Borel injections into a Borel bijection.
Cooking up specific Borel injections each way is a good exercise. But here are a couple hints in each direction:
The map $bin$ sending a real in $(0,1)$ to its canonical (= not-eventually-all-$1$s) binary representation is an injection into $omega^omega$ (in fact, into $2^omega$). $(0,1)$ and $mathbb{R}$ are clearly homeomorphic; can you show that $bin$ is Borel? (Note that we know per the above that $bin$ can't be continuous - the discontinuity crops up at the dyadic reals, do you see why?)
In the other direction, the most commonly used map is via continued fraction expansions. However, it may be more intuitive to go a bit more combinatorial. First, we can go from $omega^omega$ to $2^omega$ by "counting $0$s:" given $finomega^omega$, we write a binary sequence such that the number of $0$s between the $n$th and $(n+1)$th $1$s is $f(n)$. As an example, $f=(1,4,3,0,2,...)$ goes to $$(1,0,1,0,0,0,0,1,0,0,0,1,1,0,0,1,...)$$ This is obviously an injection, and is easily checked to be Borel; now just compose with the usual continuous map from $2^omega$ to $mathbb{R}$.
Thinking along these lines, it's also easy to check that in fact $omega^omega$ is homeomorphic to the set of irrationals (again, each with the usual topology).
Introductory texts on descriptive set theory - e.g. Kechris and Moschovakis - go into this in more detail. In particular, this is a specific example of the more general fact that any two uncountable Polish spaces are Borel isomorphic.
All of this says that the difference between the reals and the reals (hehe) is fairly minor - we can conflate the two either via bijection of fairly low complexity, or by ignoring a small (= countable) set. This means that in many situations (e.g. forcing and descriptive set theory) it essentially doesn't matter which we use.
There are situations, of course, where the difference is meaningful - e.g. in computable structure theory (here/here) - but they are in practice rare enough that the abuse of terminology doesn't lead to trouble in practice.
answered Dec 26 '18 at 18:08
Noah Schweber
121k10148284
Contra the fifth word of your question, there is no natural continuous bijection between $omega^omega$ and $mathbb{R}$. Indeed, there isn't even a continuous injection from $mathbb{R}$ to $omega^omega$ since the former is connected but the latter is totally disconnected.
However, there are still reasonably-low-complexity bijections between the two - namely, with respect to the natural topologies they are Borel isomorphic (that is, there is a Borel bijection between the two). We can see this by finding Borel injections in each direction, and then checking that the Cantor-Bernstein construction turns a pair of Borel injections into a Borel bijection.
Cooking up specific Borel injections each way is a good exercise. But here are a couple hints in each direction:
The map $bin$ sending a real in $(0,1)$ to its canonical (= not-eventually-all-$1$s) binary representation is an injection into $omega^omega$ (in fact, into $2^omega$). $(0,1)$ and $mathbb{R}$ are clearly homeomorphic; can you show that $bin$ is Borel? (Note that we know per the above that $bin$ can't be continuous - the discontinuity crops up at the dyadic reals, do you see why?)
In the other direction, the most commonly used map is via continued fraction expansions. However, it may be more intuitive to go a bit more combinatorial. First, we can go from $omega^omega$ to $2^omega$ by "counting $0$s:" given $finomega^omega$, we write a binary sequence such that the number of $0$s between the $n$th and $(n+1)$th $1$s is $f(n)$. As an example, $f=(1,4,3,0,2,...)$ goes to $$(1,0,1,0,0,0,0,1,0,0,0,1,1,0,0,1,...)$$ This is obviously an injection, and is easily checked to be Borel; now just compose with the usual continuous map from $2^omega$ to $mathbb{R}$.
Thinking along these lines, it's also easy to check that in fact $omega^omega$ is homeomorphic to the set of irrationals (again, each with the usual topology).
Introductory texts on descriptive set theory - e.g. Kechris and Moschovakis - go into this in more detail. In particular, this is a specific example of the more general fact that any two uncountable Polish spaces are Borel isomorphic.
All of this says that the difference between the reals and the reals (hehe) is fairly minor - we can conflate the two either via bijection of fairly low complexity, or by ignoring a small (= countable) set. This means that in many situations (e.g. forcing and descriptive set theory) it essentially doesn't matter which we use.
There are situations, of course, where the difference is meaningful - e.g. in computable structure theory (here/here) - but they are in practice rare enough that the abuse of terminology doesn't lead to trouble in practice.
answered Dec 26 '18 at 18:08
Noah Schweber
121k10148284
edited Dec 26 '18 at 18:26
answered Dec 26 '18 at 18:08
Noah Schweber
121k10148284
answered Dec 26 '18 at 18:08
Noah Schweber
121k10148284
answered Dec 26 '18 at 18:08
Noah Schweber
121k10148284
121k10148284
Can you explicitly construct the two Borel injections?
– user122424
Dec 26 '18 at 18:11
Yes. There are completely explicit examples of such injections.
– Andrés E. Caicedo
Dec 26 '18 at 18:14
@user122424 I've described them in a bit of detail - would you like me to flesh them out more?
– Noah Schweber
Dec 26 '18 at 18:14
@user122424 Also, note that the word "the" is unjustified - there's no unique feature of these! There are lots of reasonably-simple-to-describe Borel injections. Cooking more examples up is a good exercise, incidentally.
– Noah Schweber
Dec 26 '18 at 18:28
@user122424 Incidentally, given these questions I strongly recommend reading the first couple chapters of Kechris before continuing on with forcing - a good sense for how to "move between" essentially-equivalent spaces will help clarify things immensely.
– Noah Schweber
Dec 26 '18 at 18:38
|
show 3 more comments
Can you explicitly construct the two Borel injections?
– user122424
Dec 26 '18 at 18:11
Yes. There are completely explicit examples of such injections.
– Andrés E. Caicedo
Dec 26 '18 at 18:14
@user122424 I've described them in a bit of detail - would you like me to flesh them out more?
– Noah Schweber
Dec 26 '18 at 18:14
@user122424 Also, note that the word "the" is unjustified - there's no unique feature of these! There are lots of reasonably-simple-to-describe Borel injections. Cooking more examples up is a good exercise, incidentally.
– Noah Schweber
Dec 26 '18 at 18:28
@user122424 Incidentally, given these questions I strongly recommend reading the first couple chapters of Kechris before continuing on with forcing - a good sense for how to "move between" essentially-equivalent spaces will help clarify things immensely.
– Noah Schweber
Dec 26 '18 at 18:38
Can you explicitly construct the two Borel injections?
– user122424
Dec 26 '18 at 18:11
Can you explicitly construct the two Borel injections?
– user122424
Dec 26 '18 at 18:11
Yes. There are completely explicit examples of such injections.
– Andrés E. Caicedo
Dec 26 '18 at 18:14
Yes. There are completely explicit examples of such injections.
– Andrés E. Caicedo
Dec 26 '18 at 18:14
@user122424 I've described them in a bit of detail - would you like me to flesh them out more?
– Noah Schweber
Dec 26 '18 at 18:14
@user122424 I've described them in a bit of detail - would you like me to flesh them out more?
– Noah Schweber
Dec 26 '18 at 18:14
@user122424 Also, note that the word "the" is unjustified - there's no unique feature of these! There are lots of reasonably-simple-to-describe Borel injections. Cooking more examples up is a good exercise, incidentally.
– Noah Schweber
Dec 26 '18 at 18:28
@user122424 Also, note that the word "the" is unjustified - there's no unique feature of these! There are lots of reasonably-simple-to-describe Borel injections. Cooking more examples up is a good exercise, incidentally.
– Noah Schweber
Dec 26 '18 at 18:28
@user122424 Incidentally, given these questions I strongly recommend reading the first couple chapters of Kechris before continuing on with forcing - a good sense for how to "move between" essentially-equivalent spaces will help clarify things immensely.
– Noah Schweber
Dec 26 '18 at 18:38
@user122424 Incidentally, given these questions I strongly recommend reading the first couple chapters of Kechris before continuing on with forcing - a good sense for how to "move between" essentially-equivalent spaces will help clarify things immensely.
– Noah Schweber
Dec 26 '18 at 18:38
|
show 3 more comments
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Who calls them that?
– Hagen von Eitzen
Dec 26 '18 at 17:57
In my experience one typically identifies $omega^omega$ with the irrational elements of $mathbb R$; and then we call them "reals" because they are equinumerous, and in particular "isomorphic up to a countable set".
– Mees de Vries
Dec 26 '18 at 18:01
In books on forcing they always use reals for $omega^omega$. Am I wrong?
– user122424
Dec 26 '18 at 18:01
@HagenvonEitzen Sadly, logicians do tend to refer to elements of Baire space, elements of Cantor space, and related objects as "reals." This is because all the relevant spaces are Borel-isomorphic, so in many (but not all) contexts they are essentially interchangeable.
– Noah Schweber
Dec 26 '18 at 18:01
@HagenvonEitzen, see e.g. Wikipedia, which states "This space is commonly used in descriptive set theory, to the extent that its elements are often called “reals.”"
– Mees de Vries
Dec 26 '18 at 18:02