Are all the points in a nonempty open set limit points? My conjecture is that given any open set $A\subseteq\mathbb{R}$, all points $a\in A$ are limit points.
Prove, or if untrue, disprove by constructing a counterexample.
A few definitions for clarification:


*

*A set $A$ is open iff for every point $a\in A$, there exists a $\delta$ such that the neighborhood $\left(a-\delta,a+\delta\right)$ surrounding $a$ is completely contained within (is a subset of) $A$.

*A point $x$ is a limit point of a set $A$ iff given any $\varepsilon$, all neighborhoods $\left(x-\varepsilon,x+\varepsilon\right)$ intersect the set $A$ in some point other than $x$. Note that $x$ need not be a point in $A$.
 A: Let $a$ be a point in $A$. Since $A$ is open, there is a number $\delta > 0$ such that $(a - \delta, a + \delta) \subset A$.
We now show $a$ is a limit point of $A$. To that end, let $\epsilon > 0$ be given. We may assume, without loss of generality, that $\epsilon < \delta$. Now, $(a - \epsilon, a + \epsilon) \subset (a - \delta, a + \delta) \subset A$, so $(a - \epsilon, a + \epsilon) \cap A = (a - \epsilon, a + \epsilon)$. Since $(a - \epsilon, a + \epsilon)$ intersects $A$ in a point other than $a$, we conclude $a$ is a limit point.
A: Alternative answer
Rather than using the neighborhood definitions shown in the question, see the sequential definitions below:


*

*A set $A$ is open iff for every sequence $\left(x_n\right)\subseteq\mathbb{R}$ that converges to a point $a\in A$, $\left(x_n\right)$ is eventually in $A$. Here eventually  means there exists an $N\in\mathbb{N}$ such that when $n\gt N$, $x_n \in A$.

*A point $x$ is a limit point of a set $A$ iff there exists a sequence $\left(a_n\right)\subseteq A$, satisfying $a_n\not=x$, that converges to $x$. Note that $x$ need not be a point in $A$.

*A sequence $(b_n)$ converges to a point $b$ iff for any $\delta \gt 0$, there exists an $N\in\mathbb{N}$ such that if $n\gt N$ then $|b_n - b| < \delta$. Put another way, $b_n\in\left(b-\delta,b+\delta\right)$.
Assume $A$ is open and contains a point $a$. Because $A$ is open, every sequence $(x_n)\to a$ will eventually be contained in $A$. In other words, there exists some point in the sequence $x_N$ after which all the terms $x_n$ will be in $A$.
For any sequence $(x_n)\to a$ under the condition that $x_n\not=a$, define the sequence $(a_n)$ by
$$a_n = x_{(N+1+n)}$$
To see this clearly, notice $a_0 = x_{(N+1)}$, $a_1 = x_{(N+2)}$, etc. Essentially the sequence $(a_n)$ is the "tail end" of the sequence $(x_n)$, where the first term $a_0$ is the point at which $(x_n)$ enters the set $A$, never to leave. It is easy to see that $(a_n)\subseteq A$, $(a_n)\to a$, and $a_n\not=a$ for any $n$. Thus $a\in A$ is a limit point of $A$.
q.e.d.
A: To say that all points $a\in A$ are limit points is to say that there is no point $b\in A$ that is an isolated point.
To prove this, we will use a method called proof by contradiction, in which we assume the negation and see if we can arrive at a nonsensical conclusion. If we can arrive at a contradiction, then our assumption—namely, the negation of the conjecture—must be false. This will prove that the conjecture itself is true.
Let's assume to the contrary that an open set $A$ does indeed contain an isolated point $b$.
If $A$ is open, then every point in $A$, including $b$, must have some neighborhood that is a subset of $A$. This means that there must exist some $\delta$ such that every point within the neighborhood $\left(b-\delta,b+\delta\right)$ must be a member of the set $A$. Since the interval is open, let there be a point $d\in\left(b-\delta,b+\delta\right)$ such that $d\ne b$. Then $d\in A$.
Now, if $b$ is an isolated point, then it is not a limit point by definition. Thus we must negate the definition of limit point. If $b$ is an isolated point, there must exist some $\varepsilon$ such that the neighborhood $\left(b-\varepsilon,b+\varepsilon\right)$ intersects the set $A$ only at the point $b$. Since the interval is open, let there be a point $e\in\left(b-\varepsilon,b+\varepsilon\right)$ such that $e\ne b$. Then $e\not\in A$.
For simplicity, assign the following symbols:
$$V_\delta(b) = \left(b-\delta,b+\delta\right)$$
$$V_\varepsilon(b) = \left(b-\varepsilon,b+\varepsilon\right)$$
There are two possibilities:


*

*$$V_\delta(b) \subseteq V_\varepsilon(b)$$
Then for all $d \ne b$, we have $d\in V_\delta(b) \implies d\in V_\varepsilon(b)$. From the conclusions above, this implies $d\in A \implies d\not\in A$. This is a contradiction.

*$$V_\delta(b) \supseteq V_\varepsilon(b)$$
This is the converse of Case 1. We still arrive at a contradiction.
Since both cases lead to a contradiction, our assumption that $A$ is open and contains an isolated point $b$ is faulty. Thus there can be no such possibility. In conclusion, any open set $A$ must not contain any isolated points. In other words, all the points in a (nonempty) open set are limit points. q.e.d.

A few notes:


*

*Just because a nonempty open set contains only limit points does not mean that the set contains all its limit points. For example, all points in the open interval $(0,1)$ are limit points, but this set does not contain the limit points $0$ and $1$.

*The converse of your conjecture states, "If every point in a set is a limit point, then that set is open." This is false. The closed interval $[0,1]$ provides a counterexample: every point within this interval is a limit point, but it is not an open set.
