# Is the set $\{2, 3, 4\}$ open in some metric spaces and not open in others?

I just want to check my understanding. This is from Baby Rudin:

2.18 Definition Let $$X$$ be a metric space. All points and sets mentioned below are understood to be elements and subsets of $$X$$.

$$(a)$$ A neighborhood of $$p$$ is a set $$N_r(p)$$ consisting of all $$q$$ such that $$d(p, q) for some $$r>0$$. The number $$r$$ is called the radius of $$N_r( p)$$

$$(e)$$ A point $$p$$ is an interior point of $$E$$ if there is a neighborhood $$N$$ of $$p$$ such that $$N \subset E$$

$$(f)$$ $$E$$ is open if every point of $$E$$ is an interior point of $$E$$.

Suppose we have the metric space with set $$X=\{1, 2, 3, 4, 5\}$$ and distance function $$d(x, y)=|x-y|$$. Now $$2$$ is an interior point of $$\{2, 3, 4\}$$ because $$N_{0.5}(2)=\{2\} \subset \{2, 3, 4\}$$ (and a similar argument can be made for $$3$$ and $$4$$ as well.

But if our metric space is $$\mathbb{R}$$ with the same distance function, then $$\{2, 3, 4\}$$ is not open because no neighborhood of $$2$$ is a subset of $$\{2, 3, 4\}$$, so $$2$$ is not an interior point of $$\{2, 3, 4\}$$, right?

• Right. That's completely correct. Sep 4 '17 at 16:01
• Why "no neighborhood of $2$ is a subset of $\{2,3,4\}$"? You are right, but you need to add some explanation. Sep 4 '17 at 16:01
• @Krish When we are dealing with the set $\mathbb{R}$, every neighborhood will contain numbers which are not integers.
– Ovi
Sep 4 '17 at 16:03
• @Krish: Because any neighborhood of $2$ (by the definition given) has the form $(2-r,2+r)$ for some $r>0.$ This will readily contain some non-integer rational number. Sep 4 '17 at 16:03
• @CameronBuie sorry!!! But I was just checking whether OP understood the reason clearly or not. (+1) for the question. Sep 4 '17 at 16:07

Another thing to consider is that even when the underlying space is the same, using a different metric may yield different open sets. Letting our metric space be $\Bbb R,$ but with the distance function $$\delta(x,y):=\begin{cases}0 & x=y\\1 & x\neq y,\end{cases}$$ we can show that (for example) $\{2\}$ is a neighborhood of $2$ with radius $\frac12,$ and by similar reasoning conclude that $\{2,3,4\}$ is once again open.