Lemma 5.9.6. Let $X$ be a locally Noetherian topological space. Then $X$ is locally connected.

Proof. Let $x \in X$. Let $E$ be a neighbourhood of $x$. We have to find a connected neighbourhood of $x$ contained in $E$. By assumption there exists a neighbourhood $E'$ of $x$ which is Noetherian. Then $E \cap E'$ is Noetherian, see Lemma 5.9.2. Let $E \cap E' = Y_1 \cup \ldots \cup Y_ n$ be the decomposition into irreducible components, see Lemma 5.9.2. Let $E'' = \bigcup _{x \in Y_ i} Y_ i$. This is a connected subset of $E \cap E'$ containing $x$. It contains the open $E \cap E' \setminus (\bigcup _{x \not\in Y_ i} Y_ i)$ of $E \cap E'$ and hence it is a neighbourhood of $x$ in $X$. This proves the lemma. $\square$

There are also:

• 6 comment(s) on Section 5.9: Noetherian topological spaces

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).