From Limit Points to Derived Sets
A limit point is defined locally: every open neighborhood of the point must contain a point of the given set other than the point itself. The derived set collects all such points into a single subset of the space. This makes it possible to ask how limit points change when the original set grows, when sets are combined, or when the separation properties of the space change.
Throughout, \((X,\mathcal{T})\) is a topological space. An open neighborhood of \(x\in X\) is an open set \(U\in\mathcal{T}\) containing \(x\). A point \(x\) is a limit point of \(A\subseteq X\) when every open neighborhood of \(x\) contains a point of \(A\) different from \(x\).
The definition applies in every topological space. It does not depend on whether the space is metric, and it does not refer to sequences. The same point can be a limit point in one topology and fail to be a limit point in another, because the available open neighborhoods can differ.
Monotonicity and Finite Unions
Proof. Assume \(A\subseteq B\), and let \(x\in A'\). We must show that every open neighborhood \(U\) of \(x\) contains a point of \(B\) different from \(x\). Since \(x\in A'\), there is a point \(a\in U\cap A\) with \(a\ne x\). The inclusion \(A\subseteq B\) gives \(a\in B\), so \(U\) contains a point of \(B\) different from \(x\). This holds for every such \(U\), hence \(x\in B'\). Therefore \(A'\subseteq B'\). \(\square\)
The next result says that taking derived sets preserves finite unions exactly. The finiteness matters: for a finite union, if a point fails to be a limit point of each set, finitely many neighborhoods can be intersected to find one neighborhood witnessing failure for their union.
Proof. First consider two sets \(A,B\subseteq X\). If \(x\in A'\), every open neighborhood of \(x\) contains a point of \(A\) different from \(x\). Since \(A\subseteq A\cup B\), that neighborhood also contains a point of \(A\cup B\) different from \(x\). Thus \(x\in(A\cup B)'\). The same argument applies if \(x\in B'\), proving \(A'\cup B'\subseteq(A\cup B)'\).
For the reverse inclusion, suppose \(x\notin A'\cup B'\). Then \(x\notin A'\), so the definition gives an open neighborhood \(U\) of \(x\) containing no point of \(A\) different from \(x\). Also \(x\notin B'\), so there is an open neighborhood \(V\) of \(x\) containing no point of \(B\) different from \(x\). The intersection \(U\cap V\) is open and contains \(x\). If it contained a point of \(A\cup B\) different from \(x\), that point would belong to \(A\) or to \(B\), contradicting the defining property of \(U\) or \(V\), respectively. Hence \(x\notin(A\cup B)'\). This proves \((A\cup B)'=A'\cup B'\).
For \(n\ge 1\), apply the two-set identity repeatedly: the case \(n=1\) is immediate, and if the identity holds for \(n\), then \((\bigcup_{i=1}^{n+1}A_i)'=((\bigcup_{i=1}^{n}A_i)\cup A_{n+1})'=(\bigcup_{i=1}^{n}A_i)'\cup A_{n+1}'=\bigcup_{i=1}^{n+1}A_i'\). Finally, no point is a limit point of the empty set because no neighborhood contains a point of it. Thus \(\varnothing'=\varnothing\), as claimed for the empty family. \(\square\)
Computing Derived Sets
Worked Example: An Interval and an Isolated Point in \(\mathbb{R}\)
Use the standard topology on \(\mathbb{R}\), and let \(A=(0,1)\cup\{2\}\). We determine \(A'\) directly from neighborhoods.
If \(x\in(0,1)\), every open neighborhood of \(x\) contains an interval \((x-\varepsilon,x+\varepsilon)\) for some \(\varepsilon>0\). Choose \(\delta=\min\{\varepsilon,x,1-x\}/2\). Then \(\delta>0\), \(x+\delta\in(0,1)\), and \(x+\delta\ne x\). Thus every neighborhood contains a point of \(A\) different from \(x\), so \(x\in A'\).
At \(x=0\), any neighborhood contains \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\). The point \(y=\min\{\varepsilon,1\}/2\) satisfies \(0<y<1\) and \(y\ne0\), so \(y\in A\). Hence \(0\in A'\). At \(x=1\), the point \(y=1-\min\{\varepsilon,1\}/2\) lies in \((0,1)\) and differs from \(1\), so \(1\in A'\) as well.
The point \(2\) is not a limit point: the open neighborhood \((1.5,2.5)\) intersects \(A\) only in \(\{2\}\). If \(x<0\), the neighborhood \((x-(-x)/2,x+(-x)/2)\) lies below \(0\) and misses \(A\). If \(1<x<2\), a sufficiently small open interval centered at \(x\) lies in \((1,2)\) and misses \(A\). If \(x>2\), a sufficiently small open interval centered at \(x\) lies above \(2\) and misses \(A\). These points are not limit points.
Therefore \(A'=[0,1]\). The point \(2\) belongs to \(A\), but the neighborhood calculation shows it is not in \(A'\); membership in the original set and membership in its derived set are distinct questions.
Check: Every point of \([0,1]\) was shown to be a limit point, and every point outside \([0,1]\) was shown to have a neighborhood containing no point of \(A\) different from itself. Thus the computed set has neither omitted nor added any points.
Worked Example: A Finite Discrete Space
Let \(X=\{p,q,r\}\) with the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), so every subset is open. Take \(A=\{p,q\}\).
For \(x=p\), the open neighborhood \(\{p\}\) contains no point of \(A\) different from \(p\), so \(p\notin A'\). For \(x=q\), the neighborhood \(\{q\}\) contains no point of \(A\) different from \(q\), so \(q\notin A'\). For \(x=r\), the neighborhood \(\{r\}\) contains no point of \(A\), so \(r\notin A'\). Consequently \(A'=\varnothing\).
More generally, for every subset \(C\subseteq X\), each point \(x\in X\) has the open neighborhood \(\{x\}\). This neighborhood contains no point of \(C\) different from \(x\), so \(C'=\varnothing\), including when \(C=\varnothing\).
Check: The result \(A'=\varnothing\) agrees with monotonicity: every subset of \(A\) also has empty derived set, and no point of \(X\) is a limit point of any subset in this discrete topology.
Why the \(T_1\) Hypothesis Gives Closedness
A space is \(T_1\) if every one-point set is closed. This condition lets us remove one point from an open set without losing openness: if \(U\) is open and \(x\in X\), then \(U\setminus\{x\}=U\cap(X\setminus\{x\})\) is open. That fact is the key to proving that derived sets are closed.
Proof. We show that \(X\setminus A'\) is open. Let \(x\in X\setminus A'\). Since \(x\) is not a limit point of \(A\), there is an open neighborhood \(U\) of \(x\) such that \(U\) contains no point of \(A\) different from \(x\). We claim that \(U\cap A'=\varnothing\). The point \(x\) is not in \(A'\) by assumption. Now let \(y\in U\) with \(y\ne x\). Because \(X\) is \(T_1\), the set \(U\setminus\{x\}\) is open; it contains \(y\) and contains no point of \(A\), by the choice of \(U\). In particular it contains no point of \(A\) different from \(y\), so \(y\notin A'\). Therefore \(U\) misses \(A'\). Each \(x\in X\setminus A'\) has an open neighborhood contained in \(X\setminus A'\), proving that this complement is open and \(A'\) is closed. \(\square\)
The proof also identifies the role of \(T_1\): it ensures that a point \(y\ne x\) has a neighborhood inside \(U\) that avoids \(x\). Without that separation property, a derived set need not be closed.
Worked Example: A Derived Set That Is Not Closed
Let \(X=\{a,b\}\) have the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\), and take \(A=\{a\}\). The only open neighborhood of either \(a\) or \(b\) is \(X\). At \(a\), this neighborhood contains no point of \(A\) different from \(a\), so \(a\notin A'\). At \(b\), the neighborhood \(X\) contains \(a\in A\) with \(a\ne b\), so \(b\in A'\). Hence \(A'=\{b\}\).
The closed sets are the complements of the open sets, so they too are only \(\varnothing\) and \(X\). Thus \(\{b\}\) is not closed, and \(A'\) is not closed. This space is not \(T_1\), since the complement of \(\{a\}\) is \(\{b\}\), which is not open.
Check: The conclusion of Theorem 3 cannot be applied here because its \(T_1\) hypothesis fails. The direct neighborhood calculation gives \(A'=\{b\}\), and the list of closed sets verifies that this derived set is not closed.
A Common Pitfall
Common mistake: “A derived set is always closed.” This statement omits a necessary hypothesis. In the indiscrete example, the calculation \(A'=\{b\}\) is correct, but \(\{b\}\) is not closed. The corrected statement is that derived sets are closed in \(T_1\) spaces; without \(T_1\), closedness can fail.
A different mistake is to assume that a point of \(A\) must belong to \(A'\). In the real-line example, \(2\in A\) but \(2\notin A'\), because the neighborhood \((1.5,2.5)\) contains no other point of \(A\). The definition tests for other points of the set in every neighborhood, not merely for membership in the set.
| Question | Conclusion | Hypothesis or test |
|---|---|---|
| Does \(A\subseteq B\) imply \(A'\subseteq B'\)? | Yes | Every topological space |
| What is the derived set of a finite union? | \((A\cup B)'=A'\cup B'\) | Every topological space |
| Must \(A'\) be closed? | Yes, if \(X\) is \(T_1\) | Not guaranteed without \(T_1\) |
| Does \(x\in A\) imply \(x\in A'\)? | No | Check every neighborhood for another point of \(A\) |
Check Your Understanding
Use the definition of a limit point and state which topology is being used in each answer.
- In the standard topology on \(\mathbb{R}\), find the derived set of \(A=\{0,1\}\).
- In any topological space, prove directly that \(\varnothing'=\varnothing\).
- If \(A\subseteq B\), which inclusion between \(A'\) and \(B'\) follows, and why?
- In the indiscrete topology on \(\{a,b\}\), compute \(\{b\}'\).
- Which hypothesis in the closedness theorem is used to make \(U\setminus\{x\}\) open?