I'd like to know if my answer of the following exercise is correct. I really appreciate any suggestion you can provide to improve my argument or corrections in case I made a mistake :)
Let $a_n=1/n,\,b_m=1-1/m$ for $m,n\in \bf{N}$. Let $A_n^m$ the line segment joining $(b_m,a_n)$ with $(b_{m+1},0)$. Consider the subspace $X\subset E^2$ of the euclidean plane given by_
$$X:=\bigcup_{m\in \bf N}\bigg(\bigcup_{n\ge m} A_n^m\bigg)\cup \big([0,1]\times \{0\} \big)\cup \{(x,\sin x^{-1}):x\in [-\pi,0)\}$$
a) $X$ is connected? What are the components ?
b) For which $x\in X$, $X$ is locally connected in $x$?
c) $X$ is arcwise-connected? What are the path-components?
d) Describe the set of points $x\in X$ where $X$ is not locally connected.
a) For $m=1$, we have $A_1^1= [(0,1),(1/2,0)]$, where $[\cdot,\cdot]$ denotes line segment joining the two points, $A_2^1= [(0,1/2),(1/2,0)]$,..., $A_n^1=[(0,1/n),(1/2,0)]$. Thus
$$\bigcup_{n\ge 1} A_n^1:=\{[(0,1/n),(1/2,0)]:n\ge 1\}$$
For $m=2$, we have $A_2^2=[(1/2,1/2),(2/3,0)],\ldots, A_n^2=[(1/2,1/n),(2/3,0)]$ for $n\ge 2$. Thus
$$\bigcup_{n\ge 2} A_n^2:=\{[(1/2,1/n),(2/3,0)]:n\ge 2\}$$
Then if we call $A$ the union of all such line segment we have that form a type of infinite broom. Let $T= \{(x,\sin x^{-1}):x\in [-\pi,0)\}$. Since $T\subset T\cup \{(0,1/n)\}\subset \overline T$ (where the closure is in $E^2$), so each $T\cup \{(0,1/n)\}$ is connected. Since $T\cup \{(0,1/n)\}\cup A_n^1= T\cup A_n^1$ it follows that is connected (union of two connected spaces with common element). Then $T\cup \bigcup_{n\ge 1} A_n^1$ is connected since
$$T\cup \bigcup_{n\ge 1} A_n^1 = \bigcup_{n\ge 1} ( T\cup A_n^1)$$
is union of connected sets, namely $T\cup A_n^1$, sharing a common point $\{(1/2,0)\}$. Clearly $A\cup ([0,1]\times \{0\})$ is connected, in fact is pathwise connected. Then
$$\bigg (T\cup \bigcup_{n\ge 1} A_n^1\bigg )\cup A\cup ([0,1]\times \{0\}) =X$$
is connected. Hence $X$ is connected and only have one component, $X$ itself.
b) It is clear that for any $x\in T$, $X$ is locally connected. But it is not locally connected in the set $[0,1]\times \{0\}$ and $\{(0,1/n):n\in {\bf{N}}\}\cup \{(0,0)\}$ since every small open nhood has a separation; in one case for the infinite broom and on the other for the topologist's sine. For any other point we can find a nhood base of connected sets.
c) No, $X$ is not arcwise connected, because is not even pathwise connected. For any point in $\{(0,1/n):n\in {\bf{N}}\}\cup \{(0,0)\}$ there is not path joining it to a point in the negative x-coordinate, for example $\{(-\pi,-\sin\pi^{-1})\}$. Clearly $\overline A= A\cup ([0,1]\times \{0\})$ is path connected and similarly $T$ is path connected. Then $X$ only have two path components $\overline A$ and $T$.
d) $X$ is not locally connected, any point in $\{(0,1/n):n\in {\bf{N}}\}\cup \{(0,0)\}$ has no connected nhood. Similarly is not locally connected in $[0,1]\times \{0\}$ because for any small nhood of those points generates a separation.