Misunderstanding of the criterion to be an open subset of a metric space 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?
 A: 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 $\subseteq$.
A: In the definition of "open set", one does not require the open ball to be strictly contained in $U$. Thus that $X$ is open in $X$ is indeed trivial: for any $x \in X$, any ball $B_d(x,\epsilon)$ works, just because it is a subset of $X$.
If you understand that $\subset$ means "strictly contained" (the symbol $\subsetneq$ is more adequate instead -- usually people use $\subset$ and $\subseteq$ to mean the same thing), the conditions
(i) for every $x \in U$ there is $\epsilon > 0$ such that $B_d(x,\epsilon) \subset U$
and
(ii) for every $x \in U$ there is $\epsilon > 0$ such that $B_d(x,\epsilon) \subseteq U$
might not be equivalent. A simple counter-example is any set $X$ equipped with the discrete metric $d$. Then any singleton $\{x\}$ is open (take $0<\epsilon<1$) but there is no open ball centered in $x$ strictly contained in $\{x\}$.
