In this section we present 20 practice exercises on accumulation points, isolated points and the derived set. The exercises are arranged in order of increasing difficulty and are designed to apply, step by step, the definitions covered in the theory.
Exercise 1 — level ★☆☆☆☆
Determine the isolated points and the accumulation points of the set
\[ A=\{2,5,9\}. \]
Answer
Every point of \(A\), namely \(2,5,9\), is an isolated point. The set \(A\) has no accumulation points:
\[ A'=\varnothing. \]
Solution
The set \(A\) consists of only three points. To check that each of them is isolated, we must show that around every point there is a neighborhood containing no other element of \(A\).
Take, for instance, the point \(2\). The distance between \(2\) and its nearest neighbour in \(A\), namely \(5\), is
\[ |5-2|=3. \]
We may therefore choose, for example, the neighborhood
\[ \left(2-\frac12,2+\frac12\right). \]
This neighborhood contains \(2\), but contains neither \(5\) nor \(9\). Hence \(2\) is an isolated point.
By the same argument, \(5\) and \(9\) are isolated as well. Moreover, a finite set of real numbers has no accumulation points, since around each of its elements one can build a sufficiently small neighborhood that separates it from the remaining points of the set.
Therefore
\[ A'=\varnothing. \]
Exercise 2 — level ★☆☆☆☆
Determine the isolated points and the accumulation points of the set
\[ A=(0,1). \]
Answer
The set \(A\) has no isolated points. Its accumulation points are exactly the points of the closed interval:
\[ A'=[0,1]. \]
Solution
Every point \(x_0\in(0,1)\) is an accumulation point of \(A\). Indeed, for any \(r>0\), the neighborhood
\[ (x_0-r,x_0+r) \]
contains infinitely many points of the interval \((0,1)\) other than \(x_0\).
The point \(0\) is an accumulation point as well. In fact, for every \(r>0\), the interval
\[ (-r,r) \]
contains positive numbers less than \(1\), and hence contains elements of \(A\).
Likewise, \(1\) is an accumulation point, because every neighborhood of \(1\) contains points that are less than \(1\) and greater than \(0\).
If instead \(x_0<0\) or \(x_0>1\), we can choose a neighborhood of \(x_0\) small enough that it does not meet the interval \((0,1)\). Such points are therefore not accumulation points.
We conclude that
\[ A'=[0,1]. \]
Since every point of \(A\) is an accumulation point, no point of \(A\) is isolated.
Exercise 3 — level ★☆☆☆☆
Determine the isolated points and the derived set of
\[ A=[0,1]. \]
Answer
The set \(A\) has no isolated points and
\[ A'=[0,1]. \]
Solution
Consider first a point \(x_0\in(0,1)\). Every neighborhood of \(x_0\) contains infinitely many points of the interval \([0,1]\) other than \(x_0\), so \(x_0\) is an accumulation point.
Now consider the endpoint \(0\). Every neighborhood of \(0\), that is, every interval of the form
\[ (-r,r), \qquad r>0, \]
contains points of \([0,1]\) other than \(0\), for instance sufficiently small positive numbers. Hence \(0\) is an accumulation point.
In the same way, every neighborhood of \(1\) contains points of \([0,1]\) other than \(1\), for example numbers less than \(1\) and close enough to it. Thus \(1\) is an accumulation point too.
No point outside \([0,1]\) is an accumulation point, for if \(x_0<0\) or \(x_0>1\), there is a neighborhood of \(x_0\) that does not intersect \([0,1]\).
Therefore
\[ A'=[0,1]. \]
Since every point of \(A\) is an accumulation point, the set \(A\) has no isolated points.
Exercise 4 — level ★☆☆☆☆
Determine the isolated points and the accumulation points of
\[ A=\mathbb Z. \]
Answer
Every integer is an isolated point and
\[ A'=\varnothing. \]
Solution
Consider an arbitrary integer \(n\in\mathbb Z\). The neighborhood
\[ \left(n-\frac12,n+\frac12\right) \]
contains a single integer, namely \(n\) itself.
Indeed, the preceding integer is \(n-1\) and the next one is \(n+1\), both at distance \(1\) from \(n\). By choosing a radius smaller than \(1\), for instance \(\displaystyle \frac12\), we rule out all the other integers.
Hence every \(n\in\mathbb Z\) is an isolated point of \(\mathbb Z\).
Moreover, no real number is an accumulation point of \(\mathbb Z\). Intuitively, the integers do not cluster at any point of the real line: they are always separated from one another by a distance of \(1\).
Therefore
\[ \mathbb Z'=\varnothing. \]
Exercise 5 — level ★☆☆☆☆
Determine the derived set of
\[ A=\mathbb Q. \]
Answer
Every real number is an accumulation point of \(\mathbb Q\). Hence
\[ \mathbb Q'=\mathbb R. \]
The set \(\mathbb Q\) has no isolated points.
Solution
The key property to use is the density of the rationals in \(\mathbb R\): between any two distinct real numbers there always lies at least one rational number — in fact, infinitely many.
Let \(x_0\in\mathbb R\). We must verify that every neighborhood of \(x_0\) contains a rational number other than \(x_0\).
Consider an arbitrary neighborhood
\[ (x_0-r,x_0+r), \qquad r>0. \]
Since the rationals are dense in \(\mathbb R\), this interval contains infinitely many rational numbers. In particular, it contains at least one rational number belonging to the neighborhood and different from \(x_0\).
Hence \(x_0\) is an accumulation point of \(\mathbb Q\). As \(x_0\) was an arbitrary real number, every real number is an accumulation point of \(\mathbb Q\).
Therefore
\[ \mathbb Q'=\mathbb R. \]
Furthermore, \(\mathbb Q\) has no isolated points, since every neighborhood of a rational number contains infinitely many other rationals.
Exercise 6 — level ★★☆☆☆
Determine the isolated points and the derived set of
\[ A=\left\{\frac1n:n\in\mathbb N,\ n\ge1\right\}. \]
Answer
Every point of the form \(\displaystyle \frac1n\) is an isolated point. The only accumulation point is \(0\). Hence
\[ A'=\{0\}. \]
Solution
The elements of \(A\) are
\[ 1,\frac12,\frac13,\frac14,\ldots \]
They get closer and closer to \(0\), yet \(0\) does not belong to \(A\).
We first show that \(0\) is an accumulation point. Let \(r>0\). Since \(\displaystyle \frac1n\to0\), there exists \(n\in\mathbb N\) such that
\[ 0<\frac1n<r. \]
Thus the neighborhood \((-r,r)\) contains an element of \(A\) different from \(0\). This holds for every \(r>0\), so \(0\) is an accumulation point.
We now show that every point \(\displaystyle \frac1n\) is isolated.
If \(n=1\), it suffices to take a radius
\[ r<1-\frac12=\frac12. \]
The neighborhood \((1-r,1+r)\) contains no other element of \(A\).
If \(n\ge2\), the point \(\displaystyle \frac1n\) lies between the two consecutive terms
\[ \frac1{n-1} \qquad\text{and}\qquad \frac1{n+1}. \]
Set
\[ r=\frac12 \min\!\left\{ \frac1{n-1}-\frac1n, \frac1n-\frac1{n+1} \right\}. \]
Since both quantities inside the minimum are positive, we have \(r>0\).
With this choice the neighborhood
\[ \left(\frac1n-r,\frac1n+r\right) \]
contains no other element of \(A\). Hence \(\displaystyle \frac1n\) is an isolated point.
The only accumulation point is therefore \(0\), and so
\[ A'=\{0\}. \]
Exercise 7 — level ★★☆☆☆
Determine the isolated points and the derived set of
\[ A=\{0\}\cup\left\{\frac1n:n\in\mathbb N,\ n\ge1\right\}. \]
Answer
The point \(0\) is an accumulation point and belongs to \(A\). Every point \(\displaystyle \frac1n\) is isolated. Moreover
\[ A'=\{0\}. \]
Solution
The set consists of the point \(0\) together with the points of the sequence
\[ 1,\frac12,\frac13,\frac14,\ldots \]
The point \(0\) belongs to the set, but this does not prevent it from also being an accumulation point. Indeed, for every \(r>0\) there exists \(n\in\mathbb N\) such that
\[ 0<\frac1n<r. \]
Hence every neighborhood of \(0\) contains points of \(A\) other than \(0\).
Now consider a point of the form \(\displaystyle \frac1n\). With \(n\) fixed, this point is separated from the other elements of the set by a positive distance. We can therefore choose a sufficiently small neighborhood containing \(\displaystyle \frac1n\) alone.
Consequently, every point \(\displaystyle \frac1n\) is isolated.
The only accumulation point is \(0\), so
\[ A'=\{0\}. \]
Exercise 8 — level ★★☆☆☆
Determine the isolated points and the derived set of
\[ A=(0,1)\cup\{2\}. \]
Answer
The point \(2\) is isolated. The accumulation points are exactly the points of \([0,1]\). Hence
\[ A'=[0,1]. \]
Solution
The set \(A\) consists of the open interval \((0,1)\) together with the isolated point \(2\).
As we already know, the interval \((0,1)\) has as its accumulation points all the points of the closed interval \([0,1]\). Indeed, every neighborhood of a point of \([0,1]\) contains elements of \((0,1)\).
The point \(2\), on the other hand, is not an accumulation point. We can choose, for instance, the neighborhood
\[ \left(\frac32,\frac52\right). \]
This neighborhood contains the point \(2\), but no other element of \(A\), since the interval \((0,1)\) lies entirely to the left of \(\displaystyle \frac32\).
Hence \(2\) is an isolated point.
There are no other accumulation points: the points outside \([0,1]\) other than \(2\) can be separated from \(A\) by a suitable neighborhood, while \(2\) is isolated.
Therefore
\[ A'=[0,1]. \]
Exercise 9 — level ★★☆☆☆
Determine the isolated points and the derived set of
\[ A=[0,1]\cup\{2,3\}. \]
Answer
The points \(2\) and \(3\) are isolated. The derived set is
\[ A'=[0,1]. \]
Solution
The interval \([0,1]\) consists entirely of accumulation points. Indeed, every interior point of the interval has infinitely many points of the set arbitrarily close to it, and the same holds for the endpoints \(0\) and \(1\).
Now consider the point \(2\). We can choose a small neighborhood of \(2\), for example
\[ \left(\frac32,\frac52\right). \]
This neighborhood contains \(2\), but contains no point of \([0,1]\) and does not contain \(3\). Hence \(2\) is isolated.
Similarly, for the point \(3\) we can choose a sufficiently small neighborhood, for instance
\[ \left(\frac52,\frac72\right), \]
which contains \(3\) but no other point of \(A\). Hence \(3\) is isolated as well.
Isolated points do not belong to the derived set, since they are not accumulation points. Consequently
\[ A'=[0,1]. \]
Exercise 10 — level ★★☆☆☆
Determine the isolated points and the derived set of
\[ A=\left\{1+\frac1n:n\in\mathbb N,\ n\ge1\right\}. \]
Answer
Every point of the form \(\displaystyle 1+\frac1n\) is isolated. The only accumulation point is \(1\). Hence
\[ A'=\{1\}. \]
Solution
The elements of the set are
\[ 2,\frac32,\frac43,\frac54,\ldots \]
They are all greater than \(1\) and approach \(1\) as \(n\) grows large, because
\[ 1+\frac1n\to1. \]
Hence \(1\) is an accumulation point of \(A\). Indeed, if \(r>0\), there exists \(n\in\mathbb N\) such that
\[ \frac1n<r. \]
Then
\[ \left|\left(1+\frac1n\right)-1\right|=\frac1n<r. \]
So every neighborhood of \(1\) contains elements of \(A\).
Each point of the form \(\displaystyle 1+\frac1n\), by contrast, is isolated. Indeed, with \(n\) fixed, this point is separated from the other terms of the sequence by a positive distance.
Therefore
\[ A'=\{1\}. \]
Exercise 11 — level ★★★☆☆
Determine the isolated points and the derived set of
\[ A=(-1,0)\cup\left\{\frac1n:n\in\mathbb N,\ n\ge1\right\}. \]
Answer
The points of the form \(\displaystyle \frac1n\) are isolated. The derived set is
\[ A'=[-1,0]\cup\{0\}=[-1,0]. \]
In particular, \(0\) is an accumulation point of both the interval \((-1,0)\) and the sequence \(\displaystyle \frac1n\).
Solution
The set \(A\) is the union of two parts:
- the interval \((-1,0)\);
- the sequence \(\displaystyle 1,\frac12,\frac13,\ldots\).
The interval \((-1,0)\) has as its accumulation points all the points of the closed interval \([-1,0]\). Indeed, every interior point is plainly an accumulation point, while the endpoints \(-1\) and \(0\), although they do not belong to the interval, are approached arbitrarily closely by points of the interval.
The sequence \(\displaystyle \frac1n\) has \(0\) as its only accumulation point.
Combining the two pieces of information, we obtain
\[ A'=[-1,0]\cup\{0\}. \]
Since \(0\in[-1,0]\), this simplifies to
\[ A'=[-1,0]. \]
The points of the form \(\displaystyle \frac1n\) are isolated, because each of them can be separated from the other terms of the sequence and from the interval \((-1,0)\), which lies entirely in the negative part of the real line.
Exercise 12 — level ★★★☆☆
Determine the isolated points and the derived set of
\[ A=\left\{(-1)^n+\frac1n:n\in\mathbb N,\ n\ge1\right\}. \]
Answer
The set has two accumulation points:
\[ A'=\{-1,1\}. \]
Every element of \(A\) is an isolated point.
Solution
We study the terms of even index and those of odd index separately.
If \(n\) is even, then \((-1)^n=1\), so the corresponding terms are of the form
\[ 1+\frac1n. \]
As \(n\to\infty\), these terms tend to \(1\).
If \(n\) is odd, then \((-1)^n=-1\), so the corresponding terms are of the form
\[ -1+\frac1n. \]
As \(n\to\infty\), these terms tend to \(-1\).
The two natural candidates for accumulation points are therefore \(-1\) and \(1\).
We show that both are indeed accumulation points. Every neighborhood of \(1\) contains terms of the sequence with even index, because
\[ 1+\frac1n\to1 \]
along the even indices. Likewise, every neighborhood of \(-1\) contains terms of the sequence with odd index, because
\[ -1+\frac1n\to-1 \]
along the odd indices.
Every element of the set is isolated: once a term of the sequence is fixed, it is separated from the other terms by a positive distance, since it does not coincide with a limit point but is a single value of the sequence.
Therefore
\[ A'=\{-1,1\}. \]
Exercise 13 — level ★★★☆☆
Determine the isolated points and the derived set of
\[ A=\left\{\frac1n:n\in\mathbb N,\ n\ge1\right\}\cup\left\{2+\frac1n:n\in\mathbb N,\ n\ge1\right\}. \]
Answer
Every element of \(A\) is isolated. The accumulation points are \(0\) and \(2\), so
\[ A'=\{0,2\}. \]
Solution
The set is the union of two sequences:
\[ \frac1n\to0 \]
and
\[ 2+\frac1n\to2. \]
The first sequence has \(0\) as its only accumulation point, since its terms become arbitrarily close to \(0\).
The second sequence has \(2\) as its only accumulation point, since its terms become arbitrarily close to \(2\).
Hence certainly
\[ 0,2\in A'. \]
There are no other accumulation points. Indeed, away from \(0\) and \(2\), each sequence has only finitely many terms in any bounded region separated from these two limit points; consequently one can choose a neighborhood that contains no element of \(A\) other than the point under consideration.
Every element of the two sequences is isolated. Once a term is fixed, we can choose a neighborhood small enough to contain neither any other term of the same sequence nor any term of the other sequence.
We therefore conclude that
\[ A'=\{0,2\}. \]
Exercise 14 — level ★★★☆☆
Determine the derived set of
\[ A=\left\{\frac{n}{n+1}:n\in\mathbb N,\ n\ge1\right\}. \]
Answer
Every element of \(A\) is isolated and
\[ A'=\{1\}. \]
Solution
We rewrite the general term:
\[ \frac{n}{n+1}=1-\frac1{n+1}. \]
From this form it is immediately clear that
\[ \frac{n}{n+1}\to1. \]
Hence \(1\) is an accumulation point of \(A\). Indeed, given \(r>0\), we can choose \(n\) large enough that
\[ \frac1{n+1}<r. \]
Then
\[ \left|\frac{n}{n+1}-1\right|=\frac1{n+1}<r. \]
So every neighborhood of \(1\) contains elements of \(A\).
Every point of the form \(\displaystyle \frac{n}{n+1}\) is isolated. Indeed, the terms are distinct and, once a term is fixed, one can choose a sufficiently small neighborhood containing no other element of the sequence.
Hence the only accumulation point is \(1\):
\[ A'=\{1\}. \]
Exercise 15 — level ★★★☆☆
Determine the isolated points and the derived set of
\[ A=\left\{\frac{1}{n}:n\in\mathbb N,\ n\ge1\right\}\cup[2,3]. \]
Answer
The points of the form \(\displaystyle \frac1n\) are isolated. The derived set is
\[ A'=\{0\}\cup[2,3]. \]
Solution
The set \(A\) consists of two parts: a sequence tending to \(0\) and a closed interval \([2,3]\).
The sequence
\[ \frac1n \]
has \(0\) as its only accumulation point. All of its terms are isolated.
The interval \([2,3]\), on the other hand, has itself as its derived set. Indeed, every point of \([2,3]\), the endpoints included, is an accumulation point of the interval.
Since the sequence \(\displaystyle \frac1n\) lies in \((0,1]\) and the interval \([2,3]\) is separated from it, no further accumulation points arise between \(1\) and \(2\).
Hence the derived set is
\[ A'=\{0\}\cup[2,3]. \]
Exercise 16 — level ★★★★☆
Determine the derived set of
\[ A=\left\{\frac{m}{n}:m,n\in\mathbb N,\ 0<m<n\right\}. \]
Answer
We have
\[ A=\mathbb Q\cap(0,1), \]
and hence
\[ A'=[0,1]. \]
The set \(A\) has no isolated points.
Solution
The set \(A\) consists of all fractions \(\displaystyle \frac mn\) with \(m,n\in\mathbb N\) and \(0<m<n\). The condition \(0<m<n\) implies
\[ 0<\frac mn<1. \]
Moreover, every rational number between \(0\) and \(1\) can be written in the form \(\displaystyle \frac mn\) with \(0<m<n\). Hence
\[ A=\mathbb Q\cap(0,1). \]
Since the rationals are dense in \(\mathbb R\), every neighborhood of a point \(x_0\in(0,1)\) contains infinitely many rationals belonging to \((0,1)\). Thus every point of \((0,1)\) is an accumulation point.
The points \(0\) and \(1\) are accumulation points as well, since every neighborhood of theirs contains rationals greater than \(0\) and less than \(1\), respectively.
No point outside \([0,1]\) can be an accumulation point, since it can be separated from the interval \((0,1)\) by a suitable neighborhood.
Therefore
\[ A'=[0,1]. \]
Finally, \(A\) has no isolated points, since every neighborhood of a rational point of \((0,1)\) contains infinitely many other rationals of \((0,1)\).
Exercise 17 — level ★★★★☆
Determine the derived set of
\[ A=\mathbb Q\cap[0,1]. \]
Answer
The derived set is
\[ A'=[0,1]. \]
The set \(A\) has no isolated points.
Solution
Consider a point \(x_0\in[0,1]\). We want to show that every neighborhood of \(x_0\) contains points of \(A\) other than \(x_0\).
If \(x_0\in(0,1)\), then every neighborhood of \(x_0\) contains infinitely many rational numbers. Since the neighborhood can be chosen small enough to stay within \([0,1]\), it contains infinitely many elements of \(\mathbb Q\cap[0,1]\).
If \(x_0=0\), every neighborhood of \(0\) contains arbitrarily small positive rationals, and hence contains elements of \(A\) other than \(0\).
If \(x_0=1\), every neighborhood of \(1\) contains rationals that are less than \(1\) and arbitrarily close to it, and hence contains elements of \(A\) other than \(1\).
Thus every point of \([0,1]\) is an accumulation point of \(A\).
If instead \(x_0<0\) or \(x_0>1\), there is a neighborhood of \(x_0\) that does not intersect \([0,1]\), and hence does not intersect \(A\). Such points are not accumulation points.
We conclude that
\[ A'=[0,1]. \]
Furthermore, \(A\) has no isolated points, since every neighborhood of one of its points contains infinitely many other rationals of the interval.
Exercise 18 — level ★★★★☆
Determine whether the set
\[ A=\left\{\frac1n:n\in\mathbb N,\ n\ge1\right\} \]
is closed in \(\mathbb R\).
Answer
The set \(A\) is not closed, because
\[ A'=\{0\} \]
but \(0\notin A\).
Solution
A set \(A\subseteq\mathbb R\) is closed if and only if it contains all of its accumulation points.
In this exercise we have
\[ A=\left\{1,\frac12,\frac13,\frac14,\ldots\right\}. \]
The point \(0\) is an accumulation point of \(A\), because
\[ \frac1n\to0. \]
Indeed, for every \(r>0\) there exists \(n\in\mathbb N\) such that
\[ 0<\frac1n<r. \]
Hence every neighborhood of \(0\) contains elements of \(A\).
However, \(0\notin A\), since the elements of \(A\) are all positive and of the form \(\displaystyle \frac1n\) with \(n\ge1\).
Thus \(A\) does not contain all of its accumulation points. Therefore \(A\) is not closed in \(\mathbb R\).
Exercise 19 — level ★★★★★
Determine the accumulation points of
\[ A=\left\{\frac1n+\frac1m:n,m\in\mathbb N,\ n,m\ge1\right\}. \]
Answer
The derived set is
\[ A'=\left\{\frac1n:n\in\mathbb N,\ n\ge1\right\}\cup\{0\}. \]
Solution
The elements of \(A\) are sums of two terms of the form \(\displaystyle \frac1n\). To see where they can accumulate, we first fix one of the indices.
With \(n\) fixed, consider the sequence obtained by letting \(m\) vary:
\[ \frac1n+\frac1m. \]
Since \(\displaystyle \frac1m\to0\), we obtain
\[ \frac1n+\frac1m\to\frac1n. \]
Hence every point of the form \(\displaystyle \frac1n\) is an accumulation point of \(A\).
Furthermore, letting both indices tend to infinity, we obtain
\[ \frac1n+\frac1m\to0. \]
Hence \(0\) is an accumulation point of \(A\) as well.
We now show that there are no other accumulation points. If a sequence of distinct elements of \(A\) converges, it has the form
\[ x_k=\frac1{n_k}+\frac1{m_k}. \]
If at least one of \(n_k\) and \(m_k\) stays constant along a subsequence, then — since the elements are distinct — the other index must tend to infinity, and the possible limit is of the form \(\displaystyle \frac1n\). If instead both indices tend to infinity, then both terms tend to \(0\), and the limit is \(0\).
Hence the only accumulation points are
\[ \left\{\frac1n:n\in\mathbb N,\ n\ge1\right\}\cup\{0\}. \]
Exercise 20 — level ★★★★★
Determine the derived set of
\[ A=\left\{x\in\mathbb R:x=\frac1n+\frac{k}{m},\ n,m,k\in\mathbb N,\ 1\le k<m\right\}. \]
Answer
The derived set is
\[ A'=[0,2]. \]
Solution
Observe that the numbers of the form \(\displaystyle \frac{k}{m}\) with \(1\le k<m\) are rationals belonging to the interval \((0,1)\) and are dense in \((0,1)\).
With \(n\in\mathbb N\) fixed, the set
\[ \left\{\frac1n+\frac{k}{m}:m,k\in\mathbb N,\ 1\le k<m\right\} \]
is therefore dense in the interval
\[ \left(\frac1n,1+\frac1n\right). \]
For \(n=1\) we obtain a subset dense in \((1,2)\); for \(n=2\) we obtain a subset dense in \(\left(\frac12,\frac32\right)\); and so on.
As a consequence, every point of the interval \((0,2)\) is an accumulation point of \(A\).
The point \(0\) is an accumulation point too. Indeed, we can take \(n\) very large and, for instance, choose \(\displaystyle \frac{k}{m}\) very small. In this way we obtain elements of \(A\) that are positive and arbitrarily close to \(0\).
The point \(2\) is an accumulation point as well. Indeed, fixing \(n=1\), we can choose rationals \(\displaystyle \frac{k}{m}\in(0,1)\) arbitrarily close to \(1\). Then
\[ 1+\frac{k}{m}\to2. \]
Hence every neighborhood of \(2\) contains points of \(A\) other than \(2\).
Finally, no point outside \([0,2]\) can be an accumulation point. Indeed, every element of \(A\) satisfies
\[ 0<\frac1n+\frac{k}{m}<2. \]
Hence \(A\subseteq(0,2)\), and any point \(x_0<0\) or \(x_0>2\) can be separated from \(A\) by a suitable neighborhood.
We therefore conclude that
\[ A'=[0,2]. \]