Question

(a) This exercise will give an example of a connected space which is
not locally connected. In the plane R2
, let X0 = [0, 1] × {0},
Y0 = {0} × [0, 1], and for each n ∈ N, let Yn = {1/n} × [0,1]. Let Y = X0 ∪ (S∞
n=0 Yn). as a subspace of R
2 with its usual topology.
Prove that Y is connected but not locally connected. (Note that
this example also shows that a subspace of a locally connected
space need not be locally connected even if it is closed.)
Hint: There are many ways to prove this. One of them is as follows: Define for each n ∈ N, Zn = X0 ∪ Yn and let Z0 = X0 ∪ Y0. State why Zi is connected for each i ∈ N ∪ {0}. State why Y =S∞
i=0 union Zi is connected. To show that Y is
not locally connected, examine the point (0,1/2).
(b) Prove that an open subspace of a locally connected space is locally
connected.
(c) Consider Y = {0} ∪ {1/n: n ∈ N} as a subspace of (R , U). Use Y to show that a continuous image of a locally connected space need not be locally connected even if the function is a bijection.
Hint: The set N ∪ {0} is discrete in (R , U). A similar argument used in part
(a) could be used to show that Y is not locally connected.

Answer 1

selectat @ p.1] xfo ×e commected The poimt (o.o) inSidey (o.0) and (a ) vls a path t

1幻 0, O 벙 a path coh/ch, les inside Y. TO Show Yis mot locally damnected OeY and the open bal -L 3 2 업 Connected, open neighborhood-。

domponents. P Comes fRim tea fack that &o y is mot locally connecked x se a la eally cannacked Space, and Y be an open gulosckef X, Sincs X is lotally Conmected then there exist an

gince y is open in X and UeNts) ey is openin X hon Uin also ohen in Y aer X d Chak In () to evoy function not Comtain an Comeckeol open meyblo In