Below are some proofs involving locally connected sets. I mainly need help on finishing iii, iv, and v.
"A topological space(X,τ)is said to be locally connected if it has a basis B consisting of connected (open) sets."
(i) Verify that Z is a locally connected space which is not connected
It has been previously shown that the integers are totally disconnected, so the only connected sets are the singletons which are also open. The singleton forms a basis for Z so Z is locally connected
(ii) Show that $R^{n}$ and $S^{n} $ are locally connected, for all n>1 ($S^{2}$ is the unit circle in $R^{2}$ and $S^{3}$ is the unit sphere, ect…)
$R^{n}$ is easily proven using the first definition of locally disconnected. Every interval, or open balls are connected and open and the form a basis which were previously proven.
Take this basis from $R^{n}$ and intersect them with $S^{n}$. This forms a basis of $S^{n}$ which are all connected and open.
iii) Let (X,τ) be the subspace of $R^{2} $consisting of the points in the line segments joining〈0,1〉to 〈0,0〉and to all the points〈$\frac{1}{n}$,$0$〉, n= 1,2,3,…. Show that(X,τ)is connected but not locally connected.
We probably use contradiction to show that X is connected. Assume it is not connected. Then X=C$\cup$D where C and D are disjoint nonempty open sets in the subspace topology. I am not sure how to proceed, but I think the reason why a separation is impossible has something to do with the fact that all the line segments share (0,1) in common which makes it hard to find the disjoint sets.
As for why X is not locally connected, this time I think the problem point is (0,0). Any open neighborhood around that point intersects parts of line segment from (0,1) to ($\frac{1}{n}$,$0$);near 0.
(iv) Prove that every open subset of a locally connected space is locally connected.
Let (X,T) be a locally connected space and A open in X with the subspace topology. Then X has a set of connected basis whose intersection with A forms a basis for A. However there is no lemma or theorem proving that intersection of open connected sets with open sets are connected, so I can't conclude those basis of A are connected. How to proceed?
v) Prove the finite product topology of locally connected space is locally connected if and only if each component if locally connected.
We prove this is true for just the product of two space X and Y. If X and Y are locally connected, they each have a set of open connected basis, $B_x$ and $B_y$ We claim $B_x \times B_y $forms a connected open basis for the product topology It is previously proven that the product of connected sets are connected in the product topology. To simplify the proof, we just show that each basic open sets in the product topology are unions of elements of $B_x \times B_y $ Let $O_x \times O_y $ be an open set in X $\times$ Y. Then $O_x$=($B_{x_1} \cup$ $B_{x_2} \cup B_{x_3}$…$B_{x_n}$) and $O_y$=($B_{y_1} \cup$ $B_{y_2}\cup B_{y_3}$….$B_{y_m}$). Then $O_x \times O_y$=($B_{x_1} \cup B_{x_2}$ $\cup B_{x_3}$…$B_{x_n}$)$\times$ ($B_{y_1} \cup$ $B_{y_2} \cup B_{y_3}$….$B_{y_m}$) = $B_{x_1} \times (B_{y_1 }$ $\cup B_{y_2} \cup B_{y_3}$….$B_{y_m}$) $\cup B_{x_2} \times (B_{y_1} \cup B_{y_2} \cup B_{y_3}$….$B_{y_m}$)….We just distribute using set theory operation with Cartesian product and get $O_x \times O_y$ is the union of elements of the form $B_x \times B_y$. For finite products greater than 2, we can just use induction.
Now for the other direction, assume X $\times$ Y is locally connected so there is a set of connected open basis for X $\times$ Y. open sets are either of the form $O_x$ $\times$ $O_y$ where $O_x$ and $O_y$ are open sets from X and Y respectively or unions of sets of this form.
Question: How do I justify the fact that connected open sets in X $\times $ Y are just unions of connected open sets of the form $O_x $ $\times$ $O_y$ where $O_x$ and $O_y$ are connected open sets?
Best Answer
To show that the space in (iii) is connected, note that each of the line segments joining $\langle 0,1\rangle$ to a point on the $x$-axis is connected, and use the fact that if $\mathscr{C}$ is a family of connected sets, and there is a point $p\in\bigcap\mathscr{C}$, then $\bigcup\mathscr{C}$ is connected. If you’ve not yet proved this, now is a good time to do so. If $\bigcup\mathscr{C}$ is the union of disjoint relatively open subsets $U$ and $V$, assume without loss of generality that $p\in U$, and use the connectedness of the members of $\mathscr{C}$ to conclude that $U\supseteq C$ for each $C\in\mathscr{C}$.
You’re already on the right track to finish (iii): the space is indeed not locally connected at $\langle 0,0\rangle$. In fact, any open nbhd of $\langle 0,0\rangle$ that does not include $\langle 0,1\rangle$ fails to be connected.
For (iv) just observe that if $\mathscr{B}$ is a base for $X$, and $A\subseteq X$ is open, then $\{B\in\mathscr{B}:B\subseteq A\}$ is a base for $A$.
For the last part of (v), let $\pi_X:X\times Y\to X$ be the projection map. Let $x\in X$, and let $U$ be any open nbhd of $X$, so that $U\times Y$ is an open set in $X\times Y$. Let $y$ be any point of $Y$; $\langle x,y\rangle\in U\times Y$, so there is a connected open set $W$ in $X\times Y$ such that $\langle x,y\rangle\in W\subseteq U\times Y$. Then $\pi_X[W]$ is a connected open nbhd of $x$ contained in $U$. (Why?) It follows that $X$ has a base of connected sets and is therefore locally connected.