Lemma 5.7.11. Let $X$ be a topological space. If $X$ is locally connected, then

any open subset of $X$ is locally connected, and

the connected components of $X$ are open.

So also the connected components of open subsets of $X$ are open. In particular, every point has a fundamental system of open connected neighbourhoods.

**Proof.**
For all $x\in X$ let write $\mathcal{N}(x)$ the fundamental system of connected neighbourhoods of $x$ and let $U\subset X$ be an open subset of $X$. Then for all $x\in U$, $U$ is a neighbourhood of $x$, so the set $\{ V \in \mathcal{N}(x) | V\subset U\} $ is not empty and is a fundamental system of connected neighbourhoods of $x$ in $U$. Thus $U$ is locally connected and it proves (1).

Let $x \in \mathcal{C} \subset X$ where $\mathcal{C}$ is the connected component of $x$. Because $X$ is locally connected, there exists $\mathcal{N}$ a connected neighbourhood of $x$. Therefore by the definition of a connected component, we have $\mathcal{N} \subset \mathcal{C}$ and then $\mathcal{C}$ is a neighbourhood of $x$. It implies that $\mathcal{C}$ is a neighbourhood of each of his point, in other words $\mathcal{C}$ is open and (2) is proven.
$\square$

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