# Is the empty set a neighborhood of itself?

My topology textbook (Lee, Introduction to Manifolds) says

a $\textbf{neighborhood}$ of $p$ is just an open subset of $X$ containing $p$. More generally, if $K\subseteq X$, a neighborhood of the subset $K$ is an open subset containing $K$. (In some books, the word “neighborhood” is used in the more general sense of a subset containing an open subset containing $p$ or $K$; but for us neighborhoods are always open subsets.)

The empty set is an open subset of every topology, by definition. The empty set also "contains" itself, in the sense that $\emptyset\subseteq\emptyset$. Since it meets these two conditions, is it not a neighborhood of itself?

The answer to this related question states

The empty set ∅ is not a neighborhood of any point $x∈X$, because as you correctly observed, there are no elements of ∅.

However, nowhere in this definition are sets with no elements explicitly excluded from consideration for being neighborhoods.

• $\in$ and $\subseteq$ are not the same; thus (in mathematics) there are not two "senses" of contains. – Mauro ALLEGRANZA Jan 30 '17 at 7:29
• @MauroALLEGRANZA Am I correct in saying that $\subseteq$ is the one that mathematicians use? – Langston Jan 30 '17 at 17:04
• As often in math (alas !) terminologi is not so standard as expercted; see Element : The expressions "A includes x" and "A contains x" are also used to mean set membership, however some authors use them to mean instead "x is a subset of A". – Mauro ALLEGRANZA Jan 30 '17 at 19:14

Sure, and it matches with the idea that $\varnothing$ is a neighborhood of each of its elements (although there are no points it is a neighborhood of), because there aren't any elements not to be a neighborhood of. Every set is a neighborhood of the empty set in the sense defined there.