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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 $d(p, q)<r$ 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?

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  • $\begingroup$ Right. That's completely correct. $\endgroup$ – fleablood Sep 4 '17 at 16:01
  • $\begingroup$ Why "no neighborhood of $2$ is a subset of $\{2,3,4\}$"? You are right, but you need to add some explanation. $\endgroup$ – Krish Sep 4 '17 at 16:01
  • $\begingroup$ @Krish When we are dealing with the set $\mathbb{R}$, every neighborhood will contain numbers which are not integers. $\endgroup$ – Ovi Sep 4 '17 at 16:03
  • $\begingroup$ @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. $\endgroup$ – Cameron Buie Sep 4 '17 at 16:03
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    $\begingroup$ @CameronBuie sorry!!! But I was just checking whether OP understood the reason clearly or not. (+1) for the question. $\endgroup$ – Krish Sep 4 '17 at 16:07
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You're absolutely right. Nicely reasoned!

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.

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