Properties of connected components in locally connected spaces
Let $(X,\tau)$ be locally connected, $A \subset X$ and $C$ a connected component of $A$. Prove or give a counterexample to the following propositions:
- $C^{\circ} = C \cap A^{\circ}$
- $\delta(C) \subset \delta(A)$
- If A is a closed set then: $\delta(C) = C \cap \delta(A)$
I think 1. and 3. are false.
1.- We can take $(\mathbb{R},\tau)$ where $\tau$ is the usual topology and $A = [0,1]$. $\mathbb{R}$ is locally connected and connected, consider $x \in \mathbb{R}$ and let $C(x)$ be the connected component of $x$, since $\mathbb{R}$ is connected $C(x) = \mathbb{R}$ and since $\mathbb{R}^{\circ} = \mathbb{R}$ we can conclude that $(C(x))^{\circ} = \mathbb{R}$.
Now, if $C$ is a connected component of $A$ we will have that $C =\mathbb{R} = C^{\circ}$, therefore $C^{\circ} \neq C \cap A^{\circ} = (0,1)$.
2.- I've got the answer to this question from here and this is what I understand, but I'm not understanding everything.
By definition, the boundary of a set $Y$ is $\delta(Y) = \overline{Y} \backslash Y^{\circ} $.
Let $x \in \overline{C}$, if $x \in A^{\circ}$ (i.e. $x \in \overline{C} \cap A^{\circ}$). Since $X$ is locally connected there exist connected neighborhood $U$ such that $x \in U \subset A$. Because $C$ and $U$ are connected and $C\cap U \neq \varnothing$ we know that $C \cup U$ is connected and since $C$ is maximal $U \subset C$, therefore $x \in C^{\circ}$.
From here we will have that if $x\in \overline{C} \cap A^{\circ}$ then $x \in \overline{C} \cap C^{\circ}$
The part I'm not getting is how this implies:
$$\delta(C) = \overline{C} \backslash C^{\circ} \subset \overline{C} \backslash A^{\circ} \subset \overline{C} \backslash A^{\circ}$$
3.- Since $\delta(\mathbb{R}) = \varnothing$ we will have that $\delta(\mathbb{R}) \neq \mathbb{R} \cap \delta([0,1]) = \{0,1\}$
Solution 1:
Example 1 you give is wrong as already pointed out in the comments: $A=[0,1]$ has only one connected component, namely $C=A=[0,1]$ and for that particular case 1 does hold.
If $x \in C \cap A^\circ$ we can find an open connected $V$ so that $x \in V \subseteq A$ and by maximality of the component $C$ we have that $V \subseteq C$ (as $C \cup V$ is also a connected subset of $A$) and $x$ is an interior point of $C$. This shows the harder inclusion in $C^\circ = A^\circ \cap C$, so 1 is in fact true.
And 2 is falsified by using $A=\Bbb R$ in $X=\Bbb R$ as you did for 3.