Let $(X, d)$ be a metric space. I have been given the following definition of an open subset of $X$: A subset $U$ of $X$ is said to be open if, for every $x \in U$, there is an $r > 0$ such that $B_d(x, r) \subset U$. Now I am asked to prove $X$ is open in $X$.
At first this seemed obvious, since for any $x \in X$, $B_d(x,r) = \{y \in X: d(x,y) < r \}$ is clearly a subset of $X$. That is, if $x \in B_d(x,r)$ then $x \in X$. But then I noticed that the above definition of an open subset uses $\subset$ and not $\subseteq$, which led me to believe that I also had to show $B_d(x,r) \neq X$. I immediately ran into problems: for example, the set $Y = \{(0,1)\}$ with the Euclidean metric seems to be a metric space, but any ball centered at $(0,1)$ with radius greater than $0$ contains the entirety of $Y$. That is, for all $r > 0$ and $y \in Y$, $B_d(y, r) = Y \not \subset Y$. This led me to conclude $Y$ is not an open set in $Y$.
Have I misunderstood the definition of an open subset? Is it not "proper" inclusion but inclusion with the possibility of equality that is required?
Best Answer
In my own experience, it's very rare nowadays for the symbol $\subset$ to denote strict inclusion; I think that's because strict inclusion is not very important, so the meanings of the symbols $\subset$ and $\subseteq$ have kind of merged, as said by @Gae.S. In fact, in the rare cases where strict inclusion is actually needed, some people use $\subsetneq$.
So, feel free to understand the definition of open set where the symbol $\subset$ means the same as $\subseteq$.