Why Closing a Connected Set Cannot Split It
Connectedness rules out a division into two disjoint, nonempty open pieces. A closure adds points that every neighborhood of the original set must encounter. The key idea is that a nonempty relatively open piece of the closure cannot consist only of added points: it must meet the original set. If the closure had a separation, both pieces would therefore cut the original set, contradicting its connectedness.
We use the subspace topology throughout. If \(Y\subseteq X\), a set \(U\subseteq Y\) is open in \(Y\) when \(U=Y\cap G\) for some open set \(G\) of \(X\). A separation of \(Y\) is a pair of disjoint, nonempty sets open in \(Y\) whose union is \(Y\). A space is connected if it has no separation. A subset is connected when it is connected with its subspace topology. The empty set is connected, since it cannot be the union of two nonempty sets.
For the theorem, the relevant density is that \(A\) is dense in \(\overline{A}\). This follows from the definition of closure, but it is useful to spell out the relative-open-set version needed in the proof.
Proof. Let \(W\) be a nonempty open subset of \(Y\). By the definition of the subspace topology, \(W=Y\cap G\) for some open \(G\subseteq X\). Choose \(y\in W\). Then \(y\in G\) and \(y\in Y=\overline{A}\). Since \(G\) is an open neighborhood of \(y\), the neighborhood test for closure gives \(G\cap A\ne\varnothing\). Also \(A\subseteq Y\), so \(G\cap A=G\cap Y\cap A=W\cap A\). Thus \(W\cap A\ne\varnothing\). \(\square\)
The Closure Theorem
Proof. If \(A=\varnothing\), then \(\overline{A}=\varnothing\), which is connected. Now suppose \(A\ne\varnothing\), and put \(Y=\overline{A}\). Assume, to obtain a contradiction, that \(Y\) has a separation \(U,V\). Thus \(U,V\) are disjoint, nonempty, open in \(Y\), and \(Y=U\cup V\). By the lemma, both \(U\) and \(V\) intersect \(A\).
The sets \(U\cap A\) and \(V\cap A\) are disjoint and nonempty, and their union is \(A\). They are also open in \(A\): for example, since \(U\) is open in \(Y\), there is an open \(G\subseteq X\) with \(U=Y\cap G\), and then \(U\cap A=A\cap G\), which is open in \(A\). The same argument applies to \(V\cap A\). Consequently, \(U\cap A,V\cap A\) form a separation of \(A\), contradicting that \(A\) is connected. No separation of \(Y\) exists, so \(\overline{A}\) is connected. \(\square\)
Each part of the argument has a distinct role. Density ensures that neither side of a proposed separation can miss \(A\). The subspace topology ensures the two intersections are open in \(A\), not merely subsets of \(A\). Connectedness of \(A\) then rules out the resulting separation. The proof does not require \(A\) to be open or closed, nor does it require \(X\) itself to be connected.
Examples in Different Spaces
Worked Example: An Open Interval in the Real Line
Give \(\mathbb{R}\) its standard topology and take \(A=(0,1)\). We compute its closure and apply the theorem.
If \(x\in(0,1)\), then \(x\in A\), so \(x\in\overline{A}\). Every open neighborhood of \(0\) contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\); this interval contains \(\min(\varepsilon,1)/2\), which belongs to \((0,1)\). Hence \(0\in\overline{A}\). Every open neighborhood of \(1\) contains \((1-\varepsilon,1+\varepsilon)\) for some \(\varepsilon>0\); it contains \(1-\min(\varepsilon,1)/2\in(0,1)\), so \(1\in\overline{A}\).
If \(x<0\), the open interval \((x-|x|/2,x+|x|/2)\) is a neighborhood of \(x\) lying entirely below \(0\), so it misses \(A\). If \(x>1\), the interval \((x-(x-1)/2,x+(x-1)/2)\) lies entirely above \(1\), so it also misses \(A\). By the neighborhood test, neither such \(x\) lies in \(\overline{A}\). Therefore \(\overline{A}=[0,1]\).
The sets \((0,1)\) and \([0,1]\) are intervals, hence connected by the result on connected intervals. The theorem is consistent with this computation: closing the connected open interval adds its two endpoints without disconnecting it.
Check: The points \(0\) and \(1\) were shown to be in the closure, every point of \((0,1)\) is in the closure because it belongs to \(A\), and every point outside \([0,1]\) has a neighborhood missing \(A\). Thus the computed set is exactly \([0,1]\).
Worked Example: A Connected Singleton in the Sierpinski Space
Let \(X=\{a,b\}\) have topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\), and set \(A=\{a\}\). A singleton is connected: it cannot be expressed as the union of two disjoint nonempty subsets. We compute its closure and check connectedness of that closure directly.
The closed sets, being complements of open sets, are \(X,\{b\},\varnothing\). The only closed set containing \(\{a\}\) is \(X\). Hence \(\overline{A}=X\). To check that \(X\) is connected, consider its nonempty open subsets. They are \(\{a\}\) and \(X\). Any pair of nonempty open subsets includes \(X\), or consists of two copies of \(\{a\}\); in either case the pair cannot be disjoint and cover \(X\). Thus \(X\) has no separation.
This example also shows why relative openness matters. The closure here is the whole ambient space, but in general the pieces in a proposed separation of a closure are only guaranteed to be open in that closure. The proof correctly intersects those relatively open pieces with \(A\).
Check: The closed-set calculation gives \(\overline{\{a\}}=X\), and the open-set list verifies that \(X\) is connected. The closure is larger than the original connected set, as the theorem allows.
Worked Example: The Integers in the Finite-Complement Topology
Give \(X=\mathbb{R}\) the finite-complement topology: its open sets are \(\varnothing\) and all subsets whose complement in \(\mathbb{R}\) is finite. Let \(A=\mathbb{Z}\).
First compute the closure. Any open neighborhood \(G\) of any \(x\in\mathbb{R}\) has finite complement. Since \(\mathbb{Z}\) is infinite, it cannot be contained in that finite complement; therefore \(G\cap\mathbb{Z}\ne\varnothing\). Every \(x\in\mathbb{R}\) is consequently in \(\overline{\mathbb{Z}}\), so \(\overline{A}=\mathbb{R}\).
Now check that \(A\) is connected in its subspace topology. Its open subsets are \(\varnothing\) and the sets \(\mathbb{Z}\setminus F\) with \(F\subseteq\mathbb{Z}\) finite. Indeed, intersecting an ambient open set with \(\mathbb{Z}\) removes only finitely many integers; conversely, removing any finite \(F\) is obtained by intersecting \(\mathbb{Z}\) with the ambient open set \(\mathbb{R}\setminus F\). Any two nonempty open subsets of \(A\) have the form \(\mathbb{Z}\setminus F\) and \(\mathbb{Z}\setminus H\), where \(F,H\) are finite. Their intersection is \(\mathbb{Z}\setminus(F\cup H)\), which is nonempty because \(\mathbb{Z}\) is infinite. Therefore two nonempty open subsets of \(A\) cannot be disjoint, so \(A\) has no separation and is connected.
The same intersection argument shows that \(\mathbb{R}\) with its finite-complement topology is connected: the intersection of two nonempty open sets omits only finitely many real numbers and is nonempty. Thus both \(A\) and its closure are connected, even though the topology is not the standard topology on \(\mathbb{R}\).
Check: Each neighborhood of each real number has finite complement, while \(\mathbb{Z}\) is infinite, so every such neighborhood meets \(\mathbb{Z}\). This verifies \(\overline{\mathbb{Z}}=\mathbb{R}\) directly.
A Common Pitfall: Relative Open Does Not Mean Ambient Open
In the theorem’s proof, one must not silently treat an open subset of \(Y=\overline{A}\) as an open subset of \(X\). For a concrete example, take \(X=\mathbb{R}\) with its standard topology and \(Y=[0,1]\). The set \(U=[0,1/2)\) is open in \(Y\), since \(U=Y\cap(-1,1/2)\), but it is not open in \(\mathbb{R}\): no open interval around \(0\) is contained in \(U\).
A faulty proof might say, “\(U\) is open in \(Y\), so \(U\cap A\) is open in \(A\) because \(U\) is open in \(X\).” The claim that \(U\) is open in \(X\) is false in this example. The correction is to use the actual subspace definition: write \(U=Y\cap G\) with \(G\) open in \(X\). Since \(A\subseteq Y\), we then have \(U\cap A=G\cap A\), which is open in \(A\). This is exactly the step needed to transfer a separation of the closure to a separation of the original set.
The theorem is useful whenever connectedness is known for a set that is not closed. It lets us add all of its closure points while retaining connectedness; it says nothing, however, about whether the original set itself is closed, or whether a particular closure can be computed without examining neighborhoods.
Check Your Understanding
For each question, keep track of which topology is being used and whether openness is relative or ambient.
- State the definition of a separation of a subspace \(Y\).
- Why must each nonempty relatively open subset of \(\overline{A}\) intersect \(A\)?
- In the theorem’s proof, why are \(U\cap A\) and \(V\cap A\) open in \(A\), even when \(U\) and \(V\) are not open in \(X\)?
- In the finite-complement topology on \(\mathbb{R}\), why does every open neighborhood of a real number meet \(\mathbb{Z}\)?
- Give an example of a subset open in a subspace but not open in the ambient space, and express it as an intersection with an ambient open set.