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Hi everyone thanks for taking the time to look at this question. Consider the set $S = \{\dfrac1n \Big|\, n = 1,2,3,\cdots\}$

Here's what I understand.

$1$ is a boundary point since an open interval with $r>0$, $(1-r,1+r)$ contains points in $S$ and not in $S$.

0 is a limit point since, at the point zero regardless of how small $r$ becomes (with $r>0$) there are infinitely many points captured in our interval.

What I'm confused about.

$\dfrac12$ can't be an interior point since if we draw an open interval around it we will end of capturing points which are not a subset of $S$.

Apparently, $\dfrac12$ is an isolated point, but the definition of an isolated point is that there exists an $r>0$ such that $x$ is the only point but shouldn't there be the other points that made $\dfrac12$ not an interior point.

Apologies if any of my definitions are incorrect

Thanks again for your time.

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    What about $(1-0.001,1+0.001)$.? – 2017-02-19
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    @spaceisdarkgreen thank you thank you! It's the subtle things – 2017-02-19
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    @MyGlasses thank you for the edit and answer – 2017-02-19
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    $\frac 12$ is not interior point. Because for a point to be interior, we should be able to get an open neighborhood $U$ around $\frac 12$ such that $U \subset S$. Which is not possible here. – 2017-02-19
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    @VikrantDesai thank you – 2017-02-19
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    Isolated point. Let r= .1 then 1/3 < .4 < 1/2 < .6 < 1. So (.4,.6) is an interval for which the only point of S in (.4,.6) is 1/5. No other point in S are in the interval. So no, there shouldn't be other points and there aren't. – 2017-02-19
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    Actually, I was incorrect about boundary points. Or maybe that definition is. 1 is an isolated point and is not a limit point of S. I'll have to review the definition of boundary point to see if that fits. I thought a boundary had to be a limit point of both the set and the compliment, and not just a member of one and a limit point of the other. But I'm not sure. – 2017-02-19
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    Vikram Desai. You meant to say does not contain any element of S *other than 1/2*, I think. – 2017-02-19
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    @fleablood I meant that it is not possible to find such open neighborhood $U$ of $\frac 12$ in $\Bbb R$ which will be subset of $S$. – 2017-02-19
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    @fleablood, Every open neighborhood of $\frac 12$ in $\Bbb R$ contains points of $\Bbb R \setminus S$. Hence it can't be subset of $S$. – 2017-02-19
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    Subset of S is not nescessary. Let Q be the rationals in R. No point of Q is isolated but no point of Q contains a neighbor hood that is a subset of Q either. You'You've described any point that isn't an interior point. – 2017-02-19
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    Argh! You wrote "interior". I read "isolated" and on my phone the subset symbol wasn't displayed properly. Sorry. – 2017-02-19
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    @fleablood it's okay. No problems. :) – 2017-02-19

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$\frac{1}{2}$ is an isolated point, beauce it has two neighbours 1, and $\frac{1}{3}$ with distances $\frac{1}{2}$ and $\frac{1}{6}$ respectively. So $(\frac{1}{2}-r, \frac{1}{2}+r)$ intersects $S$ only in $\frac{1}{2}$ if $r < \frac{1}{6}$.

Because all neighbourhoods around $\frac{1}{2}$ also contain points not in $S$ it's also a boundary point of $S$, and a fortiori not an interior point.

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    Thank you very much, I didn't think about visualizing it like that with respect to the intersection. I was under the false impression that the points that made 1/2 not an interior point also made 1/2 not an isolated point. – 2017-02-19
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    This is a very clear explanation, but I balk at your use of "neighbor" to describe 1/3 and 1 in relation to 1/2. – 2017-02-19
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    @user415853 in the reals an isolated point literally lies alone, in the sense that for some distance $r$, *all* other points of $S$ are at least $r$ away. An interior point is $p$ of $S$ is one where for at least some $r$, *all* points of the space that lie closed than $r$ are also in $S$. So it's very far from being isolated. A boundary point $p$ is one where for all $r$,there are points from $S$ closer than $r$ (this could include $p$ itself if $p \in S$), as well as points not in $S$ that are closer than $r$ (could also include $p$, if $p \notin S$) – 2017-02-19
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    @fleablood they are neighbours in the induced order topology. It's a common term in linear orders: $y$ is a right hand neighbour of $x$ if $x < y$ and $(x,y) = \emptyset$ – 2017-02-19
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    Hmm, I guess it's okay. But I balk because we are assuming the ordinary topology on the reals to discuss 1/2 being isolated. "Neighbors" sounds like neighbors in the ordinary topology and I'd be inclined to think if a point has neighbors, it isn't isolated. It's probably okay, but it rubs me the wrong way. – 2017-02-19
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    @fleablood they are neighbours in $S$ and that's all that matters – 2017-02-19
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    Yeah, but I think that'd be very confusing to a student struggling with definitions. – 2017-02-19