Problem: Is set connected and compact
up vote
0
down vote
favorite
Is $ left{ (x,y, 1+x+y) in mathbb{R^3} mid x,y in [1,2] cap mathbb{I} right} $ connected as a subspace of $(mathbb{I^3} , d_{2} )$?
I guess it isn't connected since it is a subset of $mathbb{I^3}$ which isn't connected. But is it compact?
compactness connectedness
asked Nov 19 at 12:58
user560461
525
add a comment |
up vote
0
down vote
favorite
Is $ left{ (x,y, 1+x+y) in mathbb{R^3} mid x,y in [1,2] cap mathbb{I} right} $ connected as a subspace of $(mathbb{I^3} , d_{2} )$?
I guess it isn't connected since it is a subset of $mathbb{I^3}$ which isn't connected. But is it compact?
compactness connectedness
asked Nov 19 at 12:58
user560461
525
A subset of a disconnected set can be connected.
– egreg
Nov 19 at 14:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is $ left{ (x,y, 1+x+y) in mathbb{R^3} mid x,y in [1,2] cap mathbb{I} right} $ connected as a subspace of $(mathbb{I^3} , d_{2} )$?
I guess it isn't connected since it is a subset of $mathbb{I^3}$ which isn't connected. But is it compact?
compactness connectedness
asked Nov 19 at 12:58
user560461
525
Is $ left{ (x,y, 1+x+y) in mathbb{R^3} mid x,y in [1,2] cap mathbb{I} right} $ connected as a subspace of $(mathbb{I^3} , d_{2} )$?
I guess it isn't connected since it is a subset of $mathbb{I^3}$ which isn't connected. But is it compact?
compactness connectedness
compactness connectedness
asked Nov 19 at 12:58
user560461
525
asked Nov 19 at 12:58
user560461
525
edited Nov 19 at 13:04
asked Nov 19 at 12:58
user560461
525
asked Nov 19 at 12:58
user560461
525
asked Nov 19 at 12:58
user560461
525
525
A subset of a disconnected set can be connected.
– egreg
Nov 19 at 14:55
add a comment |
A subset of a disconnected set can be connected.
– egreg
Nov 19 at 14:55
A subset of a disconnected set can be connected.
– egreg
Nov 19 at 14:55
A subset of a disconnected set can be connected.
– egreg
Nov 19 at 14:55
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
I think it is not compact because the sequence $(1 + frac{1}{sqrt n},1 + frac{1}{sqrt n}, 3+ frac{2}{sqrt n})$ has a limit $(1,1,3)$ which isn't in the set.
edited Nov 19 at 14:01
GNUSupporter 8964民主女神 地下教會
12.7k72445
answered Nov 19 at 13:02
user15269
1588
I.e. it is not closed.
– Math_QED
Nov 19 at 14:32
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I think it is not compact because the sequence $(1 + frac{1}{sqrt n},1 + frac{1}{sqrt n}, 3+ frac{2}{sqrt n})$ has a limit $(1,1,3)$ which isn't in the set.
edited Nov 19 at 14:01
GNUSupporter 8964民主女神 地下教會
12.7k72445
answered Nov 19 at 13:02
user15269
1588
I.e. it is not closed.
– Math_QED
Nov 19 at 14:32
add a comment |
up vote
0
down vote
I think it is not compact because the sequence $(1 + frac{1}{sqrt n},1 + frac{1}{sqrt n}, 3+ frac{2}{sqrt n})$ has a limit $(1,1,3)$ which isn't in the set.
edited Nov 19 at 14:01
GNUSupporter 8964民主女神 地下教會
12.7k72445
answered Nov 19 at 13:02
user15269
1588
I.e. it is not closed.
– Math_QED
Nov 19 at 14:32
add a comment |
up vote
0
down vote
up vote
0
down vote
I think it is not compact because the sequence $(1 + frac{1}{sqrt n},1 + frac{1}{sqrt n}, 3+ frac{2}{sqrt n})$ has a limit $(1,1,3)$ which isn't in the set.
edited Nov 19 at 14:01
GNUSupporter 8964民主女神 地下教會
12.7k72445
answered Nov 19 at 13:02
user15269
1588
I think it is not compact because the sequence $(1 + frac{1}{sqrt n},1 + frac{1}{sqrt n}, 3+ frac{2}{sqrt n})$ has a limit $(1,1,3)$ which isn't in the set.
edited Nov 19 at 14:01
GNUSupporter 8964民主女神 地下教會
12.7k72445
answered Nov 19 at 13:02
user15269
1588
edited Nov 19 at 14:01
GNUSupporter 8964民主女神 地下教會
12.7k72445
edited Nov 19 at 14:01
GNUSupporter 8964民主女神 地下教會
12.7k72445
edited Nov 19 at 14:01
GNUSupporter 8964民主女神 地下教會
12.7k72445
12.7k72445
answered Nov 19 at 13:02
user15269
1588
answered Nov 19 at 13:02
user15269
1588
answered Nov 19 at 13:02
user15269
1588
1588
I.e. it is not closed.
– Math_QED
Nov 19 at 14:32
add a comment |
I.e. it is not closed.
– Math_QED
Nov 19 at 14:32
I.e. it is not closed.
– Math_QED
Nov 19 at 14:32
I.e. it is not closed.
– Math_QED
Nov 19 at 14:32
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004895%2fproblem-is-set-connected-and-compact%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
A subset of a disconnected set can be connected.
– egreg
Nov 19 at 14:55