I would like to prove the following:
Let $X$ be a locally connected space. Then the quasicomponents of $X$ and its components are the same.
Given a topological space $X$, the equivalence relation $\sim$ is defined as: $x \sim y$ if and only if there is no pair $A, B$ of open sets such that $x\in A, y\in B, A\cap B = \emptyset, A \cup B = X$. The quasicomponents are the equivalence classes under $\sim$. If $x \not \sim y$, they are said to be separable.
Proof:
It is pretty easy to show that every component is contained in a quasicomponent. In fact, since there can't be a separation of a component by definition, any two points in a component are also non separable.
This part of the proof doesn't require local connectedness.
Conversely, suppose $X$ is locally connected. Given a quasicomponent $Q$ and $x\in Q$, $Q$ is a neighbourhood of $x$. Thus we can find a connected neighbourhood of $x$ contained in $Q$. Clearly $Q$ is contained in the union of all such neighbourhoods for all $x$, which is a connected set and thus is contained in a component.
Is my reasoning sound? I am afraid the "union" argument might fall apart for some pathological case, plus I'm not entirely confident a quasicomponent has to contain some open set around each of its points.
Best Answer
You cannot know in advance that the union of all such connected neighrborhoods for all $x ∈ Q$ is connected. That is essentially what you trying to prove. But if you do this for a fixed $x$, then you obtain an open connected set. The point is that the components in a locally connected space are open (and so clopen).