I am not sure I understand why for the open set $(a,b)$, the limit points are $[a,b]$. Why are $a,b$ now included as limit points? Is this because we can somehow find a sequence in the open interval converging to these points?
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1Because every open set about $a$ contains points to the right which (1) live in $(a,b)$ and (2) are not equal to $a$ itself. – Randall Dec 20 '17 at 02:47
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So, we can construct a sequence of those points to the right converging to a? – user376239 Dec 20 '17 at 02:51
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2Yep. See the answer just posted below. (I prefer neighborhood definitions in topology, not sequential, but that's my taste.) – Randall Dec 20 '17 at 02:53
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The def'n of a limit point $p$ of a set $S$ is that every nbhd of $p$ contains at least one point $q$ such that $p\ne q\in S.$ – DanielWainfleet Dec 20 '17 at 03:23
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So wouldn’t that be a ,b $\in$ S – Jan 23 '21 at 20:01
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So wouldn’t that be a ,b $\in$ S because they are not included in the interval. But A is closed iff it is contains all its limit points so for our example (a,b) does not contain limit pt A. – Jan 23 '21 at 20:09
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Indeed we can find sequences that converge to those points.
Let $d = b-a$ and $\epsilon_n = \frac{d}{n}$. For each $n \in \mathbb{N}$ choose a point $x_n \in (a, a+\epsilon_n)$. Now we claim that that the sequence $\{x_n\}$ converges to $a$.
Let $\epsilon > 0$ then by construction we have
$$|x_n - a| < \epsilon_n = \frac{d}{n}$$
Thus we can let $N\in \mathbb{N}$ with $N > \frac{d}{\epsilon}$ so that for all $n \ge N$ we have
$$|x_n - a| < \epsilon_n = \frac{d}{n} < \epsilon$$
which means that $\lim x_n = a$. Thus $a$ is a limit point of $(a,b)$ and by similar approach (which I will leave to you), it is possible to show that $b$ is also a limit point.
$\boxtimes$
So what we have really done here is look at smaller and smaller intervals contained in $(a,b)$ that are close to $a$ and pick points from each of these intervals. These points form our sequence and the limit of this sequence is $a$. If you need help on constructing a sequence that converges to $b$, think about what intervals you would need to pick points from.
carsandpulsars
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So according to your definition of limit points, it sounds like the points in $\overline{A}$, if so, yes, because the points $x_{n}=a+\dfrac{b-a}{2n}$, $n=1,2,...$ belong to $(a,b)$ and are such that $x_{n}\rightarrow a$.
user284331
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If your definition is not the points in the closure, then this line of argument still works to show that every $\delta$-neighborhood about $a$ contains points of $(a,b)$ – Randall Dec 20 '17 at 02:54
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$a$ is a limit point
Sequence $x_n=a+\frac{1}{n} \in (a,b)\forall n \in \mathbb N(\because$ consequence of archemedean property). We know that $x_n \to a$ as $n \to \infty$. Hence $a$ is the limit point of $(a,b)$. ($\because$ $x\in \mathbb{R}$ is a limit point of $A$ iff there is a sequence of points $\{x_{n}\}\subset A$ which are different from $x$ and converges to $x$). Similar argument for $b$.
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What would the sequence for b look like because for a it’s straightforward but it’s not clear to me for b. – user376239 Dec 21 '17 at 23:09
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$y_n=b-\frac{1}{n}$ (perhaps not for ALL $n$, but for all $n>\frac1{b-a}$ ) – Mirko Dec 21 '17 at 23:16
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$x_n = a +\frac{1}{n} \in (a,b)$ $\forall n \in \mathbb{N}.$ I don't think this is true.consider $(a,b)=(2,2.1)$ – manifold May 28 '19 at 17:11
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It should be there is $N\ge \mathbb N$: for all $n\ge N$: $x_n=a+ a/n \in (a,b).$ Thank you for pointing out the mistake. – May 28 '19 at 17:39
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Or you can construct a decreasing sequence using the fact that $a$ is the infimum. – May 30 '19 at 02:09
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We can also try to find the closure $C$ of $(a,b)$. By definition $C$ is the intersection of all closed sets containing $(a,b)$ and $C$ will contain all limit points of $(a,b)$. The closed sets containing $(a,b)$ are of the following two forms $$A=(-\infty,d]\cup[a,+\infty)\space\space(a>d)$$ and $$B=(-\infty,b]\cup[c,+\infty)\space\space(b<c)$$ so $$C=[a,b]$$ On the other hand, by definition $C$ is the union of $(a,b)$ and the derived set $D$ of $(a,b)$ which is the set of all limit points of $(a,b)$. But $a,b\notin(a,b)$, they must come from $D$. So $a,b$ are limit points.
yushang
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There are many closed sets containing $(a,b)$ that are not of either of those two forms. For example, $[a,b] \cup K$ where $K$ is the Cantor set translated so that it doesn't intersect with $[a,b]$. – Apr 10 '20 at 04:45
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Thanks. I'll change the first part of the proof to : By definition $C$ is the smallest closed set containing $(a,b)$. $[a,b]$ is closed because its complement $(-\infty,a)\cup(b,+\infty)$ is open. The smallest sets containing $(a,b)$ are $(a,b)$,$[a,b)$,$(a,b]$ and $[a,b]$ but the former three are not closed. So $C=[a,b]$. – yushang Apr 10 '20 at 05:49
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Suppose an open interval $(c, d)$ such that it contains only $a$, we can verify it is open in Euclidean topology (Topology Without Tears, definition 2.1.1), $(c, d)$ has no intersection with $(a, b)$, so, $a$ is not a limit point of $(a, b)$. Maybe I am wrong.
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Assuming those are intervals of $\mathbb R$, the problem is that such an interval $(c,d)$ does not exist. But yes, if that interval would exist, then $a$ would not be a limit point of $(a,b)$. – celtschk Apr 02 '20 at 13:33
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Did you mean real numbers are not distinguishable from each other? Suppose $next(r)$ denote the number next to $r$, then $(r,next(next(r)))$ will contains only one real number. Maybe I am wrong. – yushang Apr 02 '20 at 16:15
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I think my problem is that real numbers are not countable, $next$ is the counting, so, it is impossible. I will learn more from link. Many thanks! – yushang Apr 02 '20 at 16:59
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Actually, the rational numbers are countable, any yet there still is no next rational number. On the other hand, there are uncountable ordinals, for which there is always a next element. This shows that countability and there existing a next number are separate concepts. Between two real numbers $x$ and $y$ there's always another real number, for example $(x+y)/2$. Thus $y$ cannot be the next number after $x$. – celtschk Apr 02 '20 at 22:08