Results in $mathbb{R}^{2n}$ applying to $mathbb{C}^{n}$
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So I managed to show that in the vectorspace $mathbb{R}^{2n}$ boundedness of a set is equivalent to total boundedness. "Intuitively" I would say that this result applies aswell to $mathbb{C}^{n}$. The problem now is that, although I know that you can view $mathbb{C}$ as a $mathbb{R}$ vectorspace, what bothers me is, what happens if we view $mathbb{C}$ as a $mathbb{C}$ vectorspace ?
And more general to what extend do results in $mathbb{R}^{2n}$ apply to $mathbb{C}^{n}$ ?
linear-algebra complex-numbers vector-spaces
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So I managed to show that in the vectorspace $mathbb{R}^{2n}$ boundedness of a set is equivalent to total boundedness. "Intuitively" I would say that this result applies aswell to $mathbb{C}^{n}$. The problem now is that, although I know that you can view $mathbb{C}$ as a $mathbb{R}$ vectorspace, what bothers me is, what happens if we view $mathbb{C}$ as a $mathbb{C}$ vectorspace ?
And more general to what extend do results in $mathbb{R}^{2n}$ apply to $mathbb{C}^{n}$ ?
linear-algebra complex-numbers vector-spaces
2
The question is quite broad, so I am not sure that it really can be answered, but at least in the example you give, the result follows from the fact that $mathbb{R}^{2n}$ and $mathbb{C}^n$ are isomorphic as metric spaces (that is, they are isometric to each other). This implies that the two spaces share any property which depends on the metric (e.g. they are homeomorphic as topological spaces). However viewing $mathbb{C}$ as a $mathbb{C}$-vector space changes the algebraic structure, and so you will start to see differences (which don't fit in a comment box).
– Xander Henderson
2 days ago
Well the first part of your answer would suffice me already! If you would turn it into an answer I would thankfully accept it.
– Christian Singer
2 days ago
add a comment |
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So I managed to show that in the vectorspace $mathbb{R}^{2n}$ boundedness of a set is equivalent to total boundedness. "Intuitively" I would say that this result applies aswell to $mathbb{C}^{n}$. The problem now is that, although I know that you can view $mathbb{C}$ as a $mathbb{R}$ vectorspace, what bothers me is, what happens if we view $mathbb{C}$ as a $mathbb{C}$ vectorspace ?
And more general to what extend do results in $mathbb{R}^{2n}$ apply to $mathbb{C}^{n}$ ?
linear-algebra complex-numbers vector-spaces
So I managed to show that in the vectorspace $mathbb{R}^{2n}$ boundedness of a set is equivalent to total boundedness. "Intuitively" I would say that this result applies aswell to $mathbb{C}^{n}$. The problem now is that, although I know that you can view $mathbb{C}$ as a $mathbb{R}$ vectorspace, what bothers me is, what happens if we view $mathbb{C}$ as a $mathbb{C}$ vectorspace ?
And more general to what extend do results in $mathbb{R}^{2n}$ apply to $mathbb{C}^{n}$ ?
linear-algebra complex-numbers vector-spaces
linear-algebra complex-numbers vector-spaces
edited 2 days ago
asked 2 days ago
Christian Singer
339113
339113
2
The question is quite broad, so I am not sure that it really can be answered, but at least in the example you give, the result follows from the fact that $mathbb{R}^{2n}$ and $mathbb{C}^n$ are isomorphic as metric spaces (that is, they are isometric to each other). This implies that the two spaces share any property which depends on the metric (e.g. they are homeomorphic as topological spaces). However viewing $mathbb{C}$ as a $mathbb{C}$-vector space changes the algebraic structure, and so you will start to see differences (which don't fit in a comment box).
– Xander Henderson
2 days ago
Well the first part of your answer would suffice me already! If you would turn it into an answer I would thankfully accept it.
– Christian Singer
2 days ago
add a comment |
2
The question is quite broad, so I am not sure that it really can be answered, but at least in the example you give, the result follows from the fact that $mathbb{R}^{2n}$ and $mathbb{C}^n$ are isomorphic as metric spaces (that is, they are isometric to each other). This implies that the two spaces share any property which depends on the metric (e.g. they are homeomorphic as topological spaces). However viewing $mathbb{C}$ as a $mathbb{C}$-vector space changes the algebraic structure, and so you will start to see differences (which don't fit in a comment box).
– Xander Henderson
2 days ago
Well the first part of your answer would suffice me already! If you would turn it into an answer I would thankfully accept it.
– Christian Singer
2 days ago
2
2
The question is quite broad, so I am not sure that it really can be answered, but at least in the example you give, the result follows from the fact that $mathbb{R}^{2n}$ and $mathbb{C}^n$ are isomorphic as metric spaces (that is, they are isometric to each other). This implies that the two spaces share any property which depends on the metric (e.g. they are homeomorphic as topological spaces). However viewing $mathbb{C}$ as a $mathbb{C}$-vector space changes the algebraic structure, and so you will start to see differences (which don't fit in a comment box).
– Xander Henderson
2 days ago
The question is quite broad, so I am not sure that it really can be answered, but at least in the example you give, the result follows from the fact that $mathbb{R}^{2n}$ and $mathbb{C}^n$ are isomorphic as metric spaces (that is, they are isometric to each other). This implies that the two spaces share any property which depends on the metric (e.g. they are homeomorphic as topological spaces). However viewing $mathbb{C}$ as a $mathbb{C}$-vector space changes the algebraic structure, and so you will start to see differences (which don't fit in a comment box).
– Xander Henderson
2 days ago
Well the first part of your answer would suffice me already! If you would turn it into an answer I would thankfully accept it.
– Christian Singer
2 days ago
Well the first part of your answer would suffice me already! If you would turn it into an answer I would thankfully accept it.
– Christian Singer
2 days ago
add a comment |
1 Answer
1
active
oldest
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up vote
1
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accepted
Your basic question could be rephrased in terms of structure. Specifically, "What structures can we impose on $mathbb{R}^{2n}$ and $mathbb{C}^n$ such that the two objects are indistinguishable? What structures would distinguish these two objects?" There are a ton of structures running around here. For example:
- As sets, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are indistinguishable: there is (at least one) bijective function between the two spaces. In particular, the map $f: mathbb{R}^2 to mathbb{C}$ given by $f(x,y) = x+iy$ is bijective, and we can go from $mathbb{R}^{2n}$ to $mathbb{C}^n$ by applying $f$ componentwise. This implies that the two sets have the same cardinality, which, from the point of view of sets, means that the two objects are the same. Therefore any set theoretic property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$.
As topological spaces, $mathbb{R}^{2n}$ and $mathbb{C}^n$. Indeed, this can be understood by seeing that as metric spaces the two objects are the same: in $mathbb{R}^{2n}$ the metric is induced by the norm
$$ | (x_1, x_2, dotsc x_{2n}) | = left( sum_{j=1}^{2n} x_j^2 right)^{1/2}, $$
and in $mathbb{C}^n$ the metric is induced by
$$ | (z_1, z_2, dotsc, z_n) |
= left( sum_{j=1}^{n} |z_j|^2 right)^{1/2}, $$
where $|z| = sqrt{Re(z)^2 + Im(z)^2}$ is the modulus of $z$. It is not too difficult to verify that the map
$$(x_1,x_2,dotsc,x_{2n-1},x_{2n}) mapsto (x_1+ix_2, dotsc, x_{2n-1}+ix_{2n})$$
is an isometry, which means that the two spaces are isometric, i.e. they are indistinguishable as metric spaces. Since the usual topologies on both spaces are induced by the metrics, this also means that the two spaces are homeomorphic as topological spaces, which means that they are indistinguishable in that context, as well.
Therefore any metric or topological property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$. As an example, you have shown that boundedness implies total boundedness in $mathbb{R}^{2n}$, which automatically implies that the same result holds for $mathbb{C}^n$. Similarly, $mathbb{R}^{2n}$ has the Heine-Borel property (a set is compact if and only if it is closed and bounded), which implies that $mathbb{C}^n$ also has this property.
Do note, however, that we could impose strange metrics or topologies on either space, and we would get something different. For example, the discrete metric on $mathbb{R}$, given by
$$ d(x,y) = begin{cases} 0 & text{if $x=y$, and} \ 1 & text{otherwise,} end{cases} $$
turns $mathbb{R}$ into a completely different space.
As vector spaces, things get a little more interesting. If we regard both $mathbb{R}^{2n}$ and $mathbb{C}^{n}$ as $mathbb{R}$-vector spaces, then they are indistinguishable. Indeed, there is an "obvious" map sending ${langle 1,0rangle, langle 0,1rangle}$ (a basis for $mathbb{R}^2$) and ${1,i}$ (a basis for $mathbb{C}$ as an $mathbb{R}$-vector space), hence the two $mathbb{R}$-vector spaces are both two dimensional vector spaces over $mathbb{R}$ and therefore isomorphic.
On the other hand, it is pretty unusual to view $mathbb{C}$ as an $mathbb{R}$-vector space–when we do this, we lose all of the interesting and useful algebraic structure which can be found in the field $mathbb{C}$. Hence it is much more common to view $mathbb{C}^n$ as a $mathbb{C}$-vector space (indeed, if we are working with vector spaces and we aren't doing this, then why would we even bother with the notation $mathbb{C}^n$?). In this setting, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are quite different spaces, owing to the differnet structures of $mathbb{C}$ and $mathbb{R}$. Hence there are things that are true about $mathbb{R}^{2n}$ which are not true about $mathbb{C}^n$, and vice versa. As a simple example, with $n=1$, $mathbb{C}$ is an algebraically closed field, while $mathbb{R}^{2}$ is not even a field.
More generally, your question touches on a branch of mathematics called category theory. I am not very conversant in category theory, so I don't want to say too much about this, but the basic goal of this theory is to study structures in mathematics in an abstract way. If we can show that two objects are isomorphic in some category, then any property which holds for one object will also hold for the other (in that category). Hence if you want to know which properties of $mathbb{R}^{2n}$ easily extend to $mathbb{C}^n$, a good place to start might be to understand the categories in which the two objects are identical, i.e. which structures can be imposed on the two objects which fail to distinguish between them.
Thanks alot for this indepth response! I appreciate that!
– Christian Singer
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your basic question could be rephrased in terms of structure. Specifically, "What structures can we impose on $mathbb{R}^{2n}$ and $mathbb{C}^n$ such that the two objects are indistinguishable? What structures would distinguish these two objects?" There are a ton of structures running around here. For example:
- As sets, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are indistinguishable: there is (at least one) bijective function between the two spaces. In particular, the map $f: mathbb{R}^2 to mathbb{C}$ given by $f(x,y) = x+iy$ is bijective, and we can go from $mathbb{R}^{2n}$ to $mathbb{C}^n$ by applying $f$ componentwise. This implies that the two sets have the same cardinality, which, from the point of view of sets, means that the two objects are the same. Therefore any set theoretic property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$.
As topological spaces, $mathbb{R}^{2n}$ and $mathbb{C}^n$. Indeed, this can be understood by seeing that as metric spaces the two objects are the same: in $mathbb{R}^{2n}$ the metric is induced by the norm
$$ | (x_1, x_2, dotsc x_{2n}) | = left( sum_{j=1}^{2n} x_j^2 right)^{1/2}, $$
and in $mathbb{C}^n$ the metric is induced by
$$ | (z_1, z_2, dotsc, z_n) |
= left( sum_{j=1}^{n} |z_j|^2 right)^{1/2}, $$
where $|z| = sqrt{Re(z)^2 + Im(z)^2}$ is the modulus of $z$. It is not too difficult to verify that the map
$$(x_1,x_2,dotsc,x_{2n-1},x_{2n}) mapsto (x_1+ix_2, dotsc, x_{2n-1}+ix_{2n})$$
is an isometry, which means that the two spaces are isometric, i.e. they are indistinguishable as metric spaces. Since the usual topologies on both spaces are induced by the metrics, this also means that the two spaces are homeomorphic as topological spaces, which means that they are indistinguishable in that context, as well.
Therefore any metric or topological property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$. As an example, you have shown that boundedness implies total boundedness in $mathbb{R}^{2n}$, which automatically implies that the same result holds for $mathbb{C}^n$. Similarly, $mathbb{R}^{2n}$ has the Heine-Borel property (a set is compact if and only if it is closed and bounded), which implies that $mathbb{C}^n$ also has this property.
Do note, however, that we could impose strange metrics or topologies on either space, and we would get something different. For example, the discrete metric on $mathbb{R}$, given by
$$ d(x,y) = begin{cases} 0 & text{if $x=y$, and} \ 1 & text{otherwise,} end{cases} $$
turns $mathbb{R}$ into a completely different space.
As vector spaces, things get a little more interesting. If we regard both $mathbb{R}^{2n}$ and $mathbb{C}^{n}$ as $mathbb{R}$-vector spaces, then they are indistinguishable. Indeed, there is an "obvious" map sending ${langle 1,0rangle, langle 0,1rangle}$ (a basis for $mathbb{R}^2$) and ${1,i}$ (a basis for $mathbb{C}$ as an $mathbb{R}$-vector space), hence the two $mathbb{R}$-vector spaces are both two dimensional vector spaces over $mathbb{R}$ and therefore isomorphic.
On the other hand, it is pretty unusual to view $mathbb{C}$ as an $mathbb{R}$-vector space–when we do this, we lose all of the interesting and useful algebraic structure which can be found in the field $mathbb{C}$. Hence it is much more common to view $mathbb{C}^n$ as a $mathbb{C}$-vector space (indeed, if we are working with vector spaces and we aren't doing this, then why would we even bother with the notation $mathbb{C}^n$?). In this setting, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are quite different spaces, owing to the differnet structures of $mathbb{C}$ and $mathbb{R}$. Hence there are things that are true about $mathbb{R}^{2n}$ which are not true about $mathbb{C}^n$, and vice versa. As a simple example, with $n=1$, $mathbb{C}$ is an algebraically closed field, while $mathbb{R}^{2}$ is not even a field.
More generally, your question touches on a branch of mathematics called category theory. I am not very conversant in category theory, so I don't want to say too much about this, but the basic goal of this theory is to study structures in mathematics in an abstract way. If we can show that two objects are isomorphic in some category, then any property which holds for one object will also hold for the other (in that category). Hence if you want to know which properties of $mathbb{R}^{2n}$ easily extend to $mathbb{C}^n$, a good place to start might be to understand the categories in which the two objects are identical, i.e. which structures can be imposed on the two objects which fail to distinguish between them.
Thanks alot for this indepth response! I appreciate that!
– Christian Singer
2 days ago
add a comment |
up vote
1
down vote
accepted
Your basic question could be rephrased in terms of structure. Specifically, "What structures can we impose on $mathbb{R}^{2n}$ and $mathbb{C}^n$ such that the two objects are indistinguishable? What structures would distinguish these two objects?" There are a ton of structures running around here. For example:
- As sets, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are indistinguishable: there is (at least one) bijective function between the two spaces. In particular, the map $f: mathbb{R}^2 to mathbb{C}$ given by $f(x,y) = x+iy$ is bijective, and we can go from $mathbb{R}^{2n}$ to $mathbb{C}^n$ by applying $f$ componentwise. This implies that the two sets have the same cardinality, which, from the point of view of sets, means that the two objects are the same. Therefore any set theoretic property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$.
As topological spaces, $mathbb{R}^{2n}$ and $mathbb{C}^n$. Indeed, this can be understood by seeing that as metric spaces the two objects are the same: in $mathbb{R}^{2n}$ the metric is induced by the norm
$$ | (x_1, x_2, dotsc x_{2n}) | = left( sum_{j=1}^{2n} x_j^2 right)^{1/2}, $$
and in $mathbb{C}^n$ the metric is induced by
$$ | (z_1, z_2, dotsc, z_n) |
= left( sum_{j=1}^{n} |z_j|^2 right)^{1/2}, $$
where $|z| = sqrt{Re(z)^2 + Im(z)^2}$ is the modulus of $z$. It is not too difficult to verify that the map
$$(x_1,x_2,dotsc,x_{2n-1},x_{2n}) mapsto (x_1+ix_2, dotsc, x_{2n-1}+ix_{2n})$$
is an isometry, which means that the two spaces are isometric, i.e. they are indistinguishable as metric spaces. Since the usual topologies on both spaces are induced by the metrics, this also means that the two spaces are homeomorphic as topological spaces, which means that they are indistinguishable in that context, as well.
Therefore any metric or topological property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$. As an example, you have shown that boundedness implies total boundedness in $mathbb{R}^{2n}$, which automatically implies that the same result holds for $mathbb{C}^n$. Similarly, $mathbb{R}^{2n}$ has the Heine-Borel property (a set is compact if and only if it is closed and bounded), which implies that $mathbb{C}^n$ also has this property.
Do note, however, that we could impose strange metrics or topologies on either space, and we would get something different. For example, the discrete metric on $mathbb{R}$, given by
$$ d(x,y) = begin{cases} 0 & text{if $x=y$, and} \ 1 & text{otherwise,} end{cases} $$
turns $mathbb{R}$ into a completely different space.
As vector spaces, things get a little more interesting. If we regard both $mathbb{R}^{2n}$ and $mathbb{C}^{n}$ as $mathbb{R}$-vector spaces, then they are indistinguishable. Indeed, there is an "obvious" map sending ${langle 1,0rangle, langle 0,1rangle}$ (a basis for $mathbb{R}^2$) and ${1,i}$ (a basis for $mathbb{C}$ as an $mathbb{R}$-vector space), hence the two $mathbb{R}$-vector spaces are both two dimensional vector spaces over $mathbb{R}$ and therefore isomorphic.
On the other hand, it is pretty unusual to view $mathbb{C}$ as an $mathbb{R}$-vector space–when we do this, we lose all of the interesting and useful algebraic structure which can be found in the field $mathbb{C}$. Hence it is much more common to view $mathbb{C}^n$ as a $mathbb{C}$-vector space (indeed, if we are working with vector spaces and we aren't doing this, then why would we even bother with the notation $mathbb{C}^n$?). In this setting, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are quite different spaces, owing to the differnet structures of $mathbb{C}$ and $mathbb{R}$. Hence there are things that are true about $mathbb{R}^{2n}$ which are not true about $mathbb{C}^n$, and vice versa. As a simple example, with $n=1$, $mathbb{C}$ is an algebraically closed field, while $mathbb{R}^{2}$ is not even a field.
More generally, your question touches on a branch of mathematics called category theory. I am not very conversant in category theory, so I don't want to say too much about this, but the basic goal of this theory is to study structures in mathematics in an abstract way. If we can show that two objects are isomorphic in some category, then any property which holds for one object will also hold for the other (in that category). Hence if you want to know which properties of $mathbb{R}^{2n}$ easily extend to $mathbb{C}^n$, a good place to start might be to understand the categories in which the two objects are identical, i.e. which structures can be imposed on the two objects which fail to distinguish between them.
Thanks alot for this indepth response! I appreciate that!
– Christian Singer
2 days ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your basic question could be rephrased in terms of structure. Specifically, "What structures can we impose on $mathbb{R}^{2n}$ and $mathbb{C}^n$ such that the two objects are indistinguishable? What structures would distinguish these two objects?" There are a ton of structures running around here. For example:
- As sets, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are indistinguishable: there is (at least one) bijective function between the two spaces. In particular, the map $f: mathbb{R}^2 to mathbb{C}$ given by $f(x,y) = x+iy$ is bijective, and we can go from $mathbb{R}^{2n}$ to $mathbb{C}^n$ by applying $f$ componentwise. This implies that the two sets have the same cardinality, which, from the point of view of sets, means that the two objects are the same. Therefore any set theoretic property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$.
As topological spaces, $mathbb{R}^{2n}$ and $mathbb{C}^n$. Indeed, this can be understood by seeing that as metric spaces the two objects are the same: in $mathbb{R}^{2n}$ the metric is induced by the norm
$$ | (x_1, x_2, dotsc x_{2n}) | = left( sum_{j=1}^{2n} x_j^2 right)^{1/2}, $$
and in $mathbb{C}^n$ the metric is induced by
$$ | (z_1, z_2, dotsc, z_n) |
= left( sum_{j=1}^{n} |z_j|^2 right)^{1/2}, $$
where $|z| = sqrt{Re(z)^2 + Im(z)^2}$ is the modulus of $z$. It is not too difficult to verify that the map
$$(x_1,x_2,dotsc,x_{2n-1},x_{2n}) mapsto (x_1+ix_2, dotsc, x_{2n-1}+ix_{2n})$$
is an isometry, which means that the two spaces are isometric, i.e. they are indistinguishable as metric spaces. Since the usual topologies on both spaces are induced by the metrics, this also means that the two spaces are homeomorphic as topological spaces, which means that they are indistinguishable in that context, as well.
Therefore any metric or topological property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$. As an example, you have shown that boundedness implies total boundedness in $mathbb{R}^{2n}$, which automatically implies that the same result holds for $mathbb{C}^n$. Similarly, $mathbb{R}^{2n}$ has the Heine-Borel property (a set is compact if and only if it is closed and bounded), which implies that $mathbb{C}^n$ also has this property.
Do note, however, that we could impose strange metrics or topologies on either space, and we would get something different. For example, the discrete metric on $mathbb{R}$, given by
$$ d(x,y) = begin{cases} 0 & text{if $x=y$, and} \ 1 & text{otherwise,} end{cases} $$
turns $mathbb{R}$ into a completely different space.
As vector spaces, things get a little more interesting. If we regard both $mathbb{R}^{2n}$ and $mathbb{C}^{n}$ as $mathbb{R}$-vector spaces, then they are indistinguishable. Indeed, there is an "obvious" map sending ${langle 1,0rangle, langle 0,1rangle}$ (a basis for $mathbb{R}^2$) and ${1,i}$ (a basis for $mathbb{C}$ as an $mathbb{R}$-vector space), hence the two $mathbb{R}$-vector spaces are both two dimensional vector spaces over $mathbb{R}$ and therefore isomorphic.
On the other hand, it is pretty unusual to view $mathbb{C}$ as an $mathbb{R}$-vector space–when we do this, we lose all of the interesting and useful algebraic structure which can be found in the field $mathbb{C}$. Hence it is much more common to view $mathbb{C}^n$ as a $mathbb{C}$-vector space (indeed, if we are working with vector spaces and we aren't doing this, then why would we even bother with the notation $mathbb{C}^n$?). In this setting, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are quite different spaces, owing to the differnet structures of $mathbb{C}$ and $mathbb{R}$. Hence there are things that are true about $mathbb{R}^{2n}$ which are not true about $mathbb{C}^n$, and vice versa. As a simple example, with $n=1$, $mathbb{C}$ is an algebraically closed field, while $mathbb{R}^{2}$ is not even a field.
More generally, your question touches on a branch of mathematics called category theory. I am not very conversant in category theory, so I don't want to say too much about this, but the basic goal of this theory is to study structures in mathematics in an abstract way. If we can show that two objects are isomorphic in some category, then any property which holds for one object will also hold for the other (in that category). Hence if you want to know which properties of $mathbb{R}^{2n}$ easily extend to $mathbb{C}^n$, a good place to start might be to understand the categories in which the two objects are identical, i.e. which structures can be imposed on the two objects which fail to distinguish between them.
Your basic question could be rephrased in terms of structure. Specifically, "What structures can we impose on $mathbb{R}^{2n}$ and $mathbb{C}^n$ such that the two objects are indistinguishable? What structures would distinguish these two objects?" There are a ton of structures running around here. For example:
- As sets, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are indistinguishable: there is (at least one) bijective function between the two spaces. In particular, the map $f: mathbb{R}^2 to mathbb{C}$ given by $f(x,y) = x+iy$ is bijective, and we can go from $mathbb{R}^{2n}$ to $mathbb{C}^n$ by applying $f$ componentwise. This implies that the two sets have the same cardinality, which, from the point of view of sets, means that the two objects are the same. Therefore any set theoretic property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$.
As topological spaces, $mathbb{R}^{2n}$ and $mathbb{C}^n$. Indeed, this can be understood by seeing that as metric spaces the two objects are the same: in $mathbb{R}^{2n}$ the metric is induced by the norm
$$ | (x_1, x_2, dotsc x_{2n}) | = left( sum_{j=1}^{2n} x_j^2 right)^{1/2}, $$
and in $mathbb{C}^n$ the metric is induced by
$$ | (z_1, z_2, dotsc, z_n) |
= left( sum_{j=1}^{n} |z_j|^2 right)^{1/2}, $$
where $|z| = sqrt{Re(z)^2 + Im(z)^2}$ is the modulus of $z$. It is not too difficult to verify that the map
$$(x_1,x_2,dotsc,x_{2n-1},x_{2n}) mapsto (x_1+ix_2, dotsc, x_{2n-1}+ix_{2n})$$
is an isometry, which means that the two spaces are isometric, i.e. they are indistinguishable as metric spaces. Since the usual topologies on both spaces are induced by the metrics, this also means that the two spaces are homeomorphic as topological spaces, which means that they are indistinguishable in that context, as well.
Therefore any metric or topological property which holds for $mathbb{R}^{2n}$ must also hold for $mathbb{C}^n$. As an example, you have shown that boundedness implies total boundedness in $mathbb{R}^{2n}$, which automatically implies that the same result holds for $mathbb{C}^n$. Similarly, $mathbb{R}^{2n}$ has the Heine-Borel property (a set is compact if and only if it is closed and bounded), which implies that $mathbb{C}^n$ also has this property.
Do note, however, that we could impose strange metrics or topologies on either space, and we would get something different. For example, the discrete metric on $mathbb{R}$, given by
$$ d(x,y) = begin{cases} 0 & text{if $x=y$, and} \ 1 & text{otherwise,} end{cases} $$
turns $mathbb{R}$ into a completely different space.
As vector spaces, things get a little more interesting. If we regard both $mathbb{R}^{2n}$ and $mathbb{C}^{n}$ as $mathbb{R}$-vector spaces, then they are indistinguishable. Indeed, there is an "obvious" map sending ${langle 1,0rangle, langle 0,1rangle}$ (a basis for $mathbb{R}^2$) and ${1,i}$ (a basis for $mathbb{C}$ as an $mathbb{R}$-vector space), hence the two $mathbb{R}$-vector spaces are both two dimensional vector spaces over $mathbb{R}$ and therefore isomorphic.
On the other hand, it is pretty unusual to view $mathbb{C}$ as an $mathbb{R}$-vector space–when we do this, we lose all of the interesting and useful algebraic structure which can be found in the field $mathbb{C}$. Hence it is much more common to view $mathbb{C}^n$ as a $mathbb{C}$-vector space (indeed, if we are working with vector spaces and we aren't doing this, then why would we even bother with the notation $mathbb{C}^n$?). In this setting, $mathbb{R}^{2n}$ and $mathbb{C}^n$ are quite different spaces, owing to the differnet structures of $mathbb{C}$ and $mathbb{R}$. Hence there are things that are true about $mathbb{R}^{2n}$ which are not true about $mathbb{C}^n$, and vice versa. As a simple example, with $n=1$, $mathbb{C}$ is an algebraically closed field, while $mathbb{R}^{2}$ is not even a field.
More generally, your question touches on a branch of mathematics called category theory. I am not very conversant in category theory, so I don't want to say too much about this, but the basic goal of this theory is to study structures in mathematics in an abstract way. If we can show that two objects are isomorphic in some category, then any property which holds for one object will also hold for the other (in that category). Hence if you want to know which properties of $mathbb{R}^{2n}$ easily extend to $mathbb{C}^n$, a good place to start might be to understand the categories in which the two objects are identical, i.e. which structures can be imposed on the two objects which fail to distinguish between them.
answered 2 days ago
Xander Henderson
13.8k93552
13.8k93552
Thanks alot for this indepth response! I appreciate that!
– Christian Singer
2 days ago
add a comment |
Thanks alot for this indepth response! I appreciate that!
– Christian Singer
2 days ago
Thanks alot for this indepth response! I appreciate that!
– Christian Singer
2 days ago
Thanks alot for this indepth response! I appreciate that!
– Christian Singer
2 days ago
add a comment |
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The question is quite broad, so I am not sure that it really can be answered, but at least in the example you give, the result follows from the fact that $mathbb{R}^{2n}$ and $mathbb{C}^n$ are isomorphic as metric spaces (that is, they are isometric to each other). This implies that the two spaces share any property which depends on the metric (e.g. they are homeomorphic as topological spaces). However viewing $mathbb{C}$ as a $mathbb{C}$-vector space changes the algebraic structure, and so you will start to see differences (which don't fit in a comment box).
– Xander Henderson
2 days ago
Well the first part of your answer would suffice me already! If you would turn it into an answer I would thankfully accept it.
– Christian Singer
2 days ago