# 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?

• It should be $\subseteq$, and I wouldn't be surprised if the author is adopting the notation "$\subset$" = "$\subseteq$".
– user239203
Commented Jan 10, 2020 at 3:33
• What if $r=\frac12?$ Commented Jan 10, 2020 at 3:35

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 $$\subseteq$$.

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 $$\{x\}$$.

• Great, thanks! Yes for some reason I have always understood $\subset$ to mean strict inclusion, but I see now that it is more commonly used interchangeably with $\subseteq$.
– bxs
Commented Jan 10, 2020 at 3:41
• Personally I don't like $\subset$ just because it may lead to this confusion. I only deal in extremes: $\subseteq$ or $\subsetneq$ if I need to emphasize anything. Commented Jan 10, 2020 at 3:47