One-Dimensional Subsets of Hilbert Space vs. Euclidean Space
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Motivated by a question I received (though a bit different), is there a one-dimensional subset of Hilbert Space that cannot be embedded in $mathbbR^3$?
Here, dimension refers to covering or inductive dimension. I know by the Menger-Nobeling Theorem that any compact, one-dimensional subset of Hilbert Space can be embedded in three-space, but compactifications of Hilbert Space are not something I've studied.
It feels like it might be equivalent to ask if the Hilbert Space and Hilbert Cube can be distinguished by their one-dimensional subsets, as well. Curious to know the answer to either! Thanks!
general-topology hilbert-spaces dimension-theory
add a comment |Â
up vote
1
down vote
favorite
Motivated by a question I received (though a bit different), is there a one-dimensional subset of Hilbert Space that cannot be embedded in $mathbbR^3$?
Here, dimension refers to covering or inductive dimension. I know by the Menger-Nobeling Theorem that any compact, one-dimensional subset of Hilbert Space can be embedded in three-space, but compactifications of Hilbert Space are not something I've studied.
It feels like it might be equivalent to ask if the Hilbert Space and Hilbert Cube can be distinguished by their one-dimensional subsets, as well. Curious to know the answer to either! Thanks!
general-topology hilbert-spaces dimension-theory
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Motivated by a question I received (though a bit different), is there a one-dimensional subset of Hilbert Space that cannot be embedded in $mathbbR^3$?
Here, dimension refers to covering or inductive dimension. I know by the Menger-Nobeling Theorem that any compact, one-dimensional subset of Hilbert Space can be embedded in three-space, but compactifications of Hilbert Space are not something I've studied.
It feels like it might be equivalent to ask if the Hilbert Space and Hilbert Cube can be distinguished by their one-dimensional subsets, as well. Curious to know the answer to either! Thanks!
general-topology hilbert-spaces dimension-theory
Motivated by a question I received (though a bit different), is there a one-dimensional subset of Hilbert Space that cannot be embedded in $mathbbR^3$?
Here, dimension refers to covering or inductive dimension. I know by the Menger-Nobeling Theorem that any compact, one-dimensional subset of Hilbert Space can be embedded in three-space, but compactifications of Hilbert Space are not something I've studied.
It feels like it might be equivalent to ask if the Hilbert Space and Hilbert Cube can be distinguished by their one-dimensional subsets, as well. Curious to know the answer to either! Thanks!
general-topology hilbert-spaces dimension-theory
asked Aug 22 at 7:40
John Samples
1,031516
1,031516
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The answer is no. It is a generalization of the above-stated Menger-Nobeling Theorem, given as theorem V.3 in Hurewicz/Wallman, that the assumption of compactness may be dropped. In other words, any separable metric space of dimension $n$ can be embedded in $mathbbR^2n+1$.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The answer is no. It is a generalization of the above-stated Menger-Nobeling Theorem, given as theorem V.3 in Hurewicz/Wallman, that the assumption of compactness may be dropped. In other words, any separable metric space of dimension $n$ can be embedded in $mathbbR^2n+1$.
add a comment |Â
up vote
1
down vote
accepted
The answer is no. It is a generalization of the above-stated Menger-Nobeling Theorem, given as theorem V.3 in Hurewicz/Wallman, that the assumption of compactness may be dropped. In other words, any separable metric space of dimension $n$ can be embedded in $mathbbR^2n+1$.
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The answer is no. It is a generalization of the above-stated Menger-Nobeling Theorem, given as theorem V.3 in Hurewicz/Wallman, that the assumption of compactness may be dropped. In other words, any separable metric space of dimension $n$ can be embedded in $mathbbR^2n+1$.
The answer is no. It is a generalization of the above-stated Menger-Nobeling Theorem, given as theorem V.3 in Hurewicz/Wallman, that the assumption of compactness may be dropped. In other words, any separable metric space of dimension $n$ can be embedded in $mathbbR^2n+1$.
answered Aug 22 at 9:11
John Samples
1,031516
1,031516
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2890734%2fone-dimensional-subsets-of-hilbert-space-vs-euclidean-space%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password