Lemma 5.21.4. Let $X$ be a topological space. Let $X = \bigcup U_ i$ be an open covering. Let $T \subset X$ be a subset. If $T \cap U_ i$ is nowhere dense in $U_ i$ for all $i$, then $T$ is nowhere dense in $X$.
Proof. Denote $\overline{T}_ i$ the closure of $T \cap U_ i$ in $U_ i$. We have $\overline{T} \cap U_ i = \overline{T}_ i$. Taking the interior commutes with intersection with opens, thus
\[ (\text{interior of }\overline{T}) \cap U_ i = \text{interior of }(\overline{T} \cap U_ i) = \text{interior in }U_ i\text{ of }\overline{T}_ i \]
By assumption the last of these is empty. Hence $T$ is nowhere dense in $X$. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: