Problem: Is set connected and compact











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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







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  • A subset of a disconnected set can be connected.
    – egreg
    yesterday















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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







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  • A subset of a disconnected set can be connected.
    – egreg
    yesterday













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







share|cite|improve this question















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






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share|cite|improve this question













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share|cite|improve this question








edited yesterday

























asked yesterday









user560461

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525












  • A subset of a disconnected set can be connected.
    – egreg
    yesterday


















  • A subset of a disconnected set can be connected.
    – egreg
    yesterday
















A subset of a disconnected set can be connected.
– egreg
yesterday




A subset of a disconnected set can be connected.
– egreg
yesterday










1 Answer
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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.






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  • I.e. it is not closed.
    – Math_QED
    yesterday











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1 Answer
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1 Answer
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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.






share|cite|improve this answer























  • I.e. it is not closed.
    – Math_QED
    yesterday















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.






share|cite|improve this answer























  • I.e. it is not closed.
    – Math_QED
    yesterday













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.






share|cite|improve this answer














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.







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edited yesterday









GNUSupporter 8964民主女神 地下教會

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answered yesterday









user15269

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  • I.e. it is not closed.
    – Math_QED
    yesterday


















  • I.e. it is not closed.
    – Math_QED
    yesterday
















I.e. it is not closed.
– Math_QED
yesterday




I.e. it is not closed.
– Math_QED
yesterday


















 

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