Let subspace A of X be path-connected
equivalence relation x~y is defined as there exists a path from x to y
fix a x in A
for any y in A, x~y
then there exists a path f : [a,b(y)] -> A by f(a)=x, f(b)=y
since [a,b] is connected and f is continuous, f([a,b]) is connected
[intersects of f([a,b(y)]) for any y in A] contains f(a) = x
i.e. it's not empty
thus union of f([a,b(y)]) for any y in A is connected
이렇게 떠올려서 했는데
맞는지 모르겠네요
아니면 다른 증명이 있을까요?
equivalence relation x~y is defined as there exists a path from x to y
fix a x in A
for any y in A, x~y
then there exists a path f : [a,b(y)] -> A by f(a)=x, f(b)=y
since [a,b] is connected and f is continuous, f([a,b]) is connected
[intersects of f([a,b(y)]) for any y in A] contains f(a) = x
i.e. it's not empty
thus union of f([a,b(y)]) for any y in A is connected
이렇게 떠올려서 했는데
맞는지 모르겠네요
아니면 다른 증명이 있을까요?
해당 댓글은 삭제되었습니다.
아. 넵. 첨에 떠올릴땐 설마 paste lem써야 하나 싶어서 a ,b(y)라 했습니다. 그런데 a, b말고 0, 1로 해도 웬만하면 무방하죠?