Definition 18.10.
A subset \(S\) of a connected topological space \(X\) is a cut set of \(X\) if the set \(X \setminus S\) is disconnected. A point \(p\) in a connected topological space \(X\) is a cut point if \(X \setminus \{p\}\) is disconnected.
Section Cut Sets
It can be difficult to determine if two topological spaces are homeomorphic. We can sometimes use topological invariants to determine if spaces are not homeomorphic. For example, if \(X\) is connected and \(Y\) is not, then \(X\) and \(Y\) are not homeomorphic. But just because two spaces are connected, it does not automatically follow that the spaces are homeomorphic. For example consider the spaces \((0,2)\) and \([0,2)\text{.}\) Both are connected subsets of \(\R\text{.}\) If we remove a point, say 1, from the set \((0,2)\) the resulting space \((0,1) \cup (1,2)\) is no longer connected. The same result is true if we remove any point from \((0,2)\text{.}\) However, if we remove the point \(0\) from \([0,2)\) the resulting space \((0,2)\) is connected. So the spaces \((0,2)\) and \([0,2)\) are fundamentally different in this respect, and so are not homeomorphic. Any set that we can remove from a connected set to obtain a disconnected set is called a cut set.
Example 18.11.
(a)
The point \(1\) is a cut point of the space \((0,2)\text{.}\) In fact, every point in \((0,2)\) is a cut point of \((0,2)\text{.}\)
(b)
Let \(D = \{(x,y) \mid x^2+y^2 \leq 4\}\) in \((\R^2, d_E)\text{.}\) That is, \(D\) is the closed disk of radius \(2\) in the plane. The set \(D\) has no cut points. However, if \(S = \{(x,y) \mid x^2+y^2 = 1\}\text{,}\) then \(D \setminus S\) consists of two connected components: the open ball \(B((0,0),1)\) and the annulus \(\{(x,y) \mid 1 \lt x^2+y^2 \leq 4\}\) as illustrated in Figure 18.12. So \(S\) is a cut set of \(D\text{.}\)
Once we have a new property, we then ask if that property is a topological invariant.
Theorem 18.13.
Let \(X\) and \(Y\) be connected topological spaces and let \(f:X \to Y\) be a homeomorphism. If \(S \subset X\) is a cut set, then \(f(S)\) is a cut set of \(Y\text{.}\)
Proof.
Let \(X\) and \(Y\) be topological spaces with \(f : X \to Y\) a homeomorphism. Let \(S\) be a cut set of \(X\text{.}\) Let \(U\) and \(V\) form a separation of \(X \setminus S\text{.}\) We will demonstrate that \(f(U)\) and \(f(V)\) form a separation of \(Y \setminus f(S)\text{,}\) which will prove that \(f(S)\) is a cut set of \(Y\text{.}\) Since \(f^{-1}\) is continuous, the sets \(f(U)\) and \(f(V)\) are open sets in \(Y\text{.}\) Next we prove that \((Y \setminus f(S)) \subseteq (f(U) \cup f(V))\text{.}\) Let \(y \in Y \setminus f(S)\text{.}\) Since \(f\) is a surjection, there exists an \(x \in X\) with \(f(x) = y\text{.}\) The fact that \(y \notin f(S)\) means that \(x \notin S\text{.}\) So \(x \in (X \setminus S) \subseteq (U \cup V)\text{.}\) If \(x \in U\text{,}\) then \(f(x) = y \in f(U)\) and if \(x \in V\text{,}\) then \(x = f(y) \in f(V)\text{.}\) So \((Y \setminus f(S)) \subseteq (f(U) \cup f(V))\text{.}\)
Now we demonstrate that \(f(U) \cap (Y \setminus f(S)) \neq \emptyset\) and \(f(V) \cap (Y \setminus f(S)) \neq \emptyset\text{.}\) Since \(U\) and \(V\) form a separation of \(X \setminus S\text{,}\) we know that \(U \cap (X \setminus S) \neq \emptyset\) and \(V \cap (X \setminus S) \neq \emptyset\text{.}\) Let \(x \in U \cap (X \setminus S)\text{.}\) Then \(x \in U\) and \(x \notin S\text{.}\) So \(f(x) \in f(U)\) and the fact that \(f\) is an injection implies that \(f(x) \notin f(S)\text{.}\) Thus, \(f(x) \in f(U) \cap (Y \setminus f(S))\text{.}\) The same argument shows that \(x \in V \cap (X \setminus S)\) implies that \(f(x) \in f(V) \cap (Y \setminus f(S))\text{.}\) So \(f(U) \cap (Y \setminus f(S)) \neq \emptyset\) and \(f(V) \cap (Y \setminus f(S)) \neq \emptyset\text{.}\)
Finally, we show that \(f(U) \cap f(V) \cap (Y \setminus f(S)) = \emptyset\text{.}\) Suppose \(y \in f(U) \cap f(V) \cap (Y \setminus f(S))\text{.}\) Let \(x \in X\) such that \(f(x) = y\text{.}\) Since \(f\) is an injection, we know that \(f(x) \in f(U)\) means \(x \in U\text{.}\) so \(x \in U \cap V\text{.}\) The fact that \(y \in Y \setminus f(S)\) means that \(y \notin f(S)\text{.}\) Thus, \(x \notin S\text{.}\) So \(x \in X \setminus S\text{.}\) We then have \(x \in U \cap V \cap (X \setminus S) = \emptyset\text{.}\) It follows that \(f(U) \cap f(V) \cap (Y \setminus f(S)) = \emptyset\text{.}\) Therefore, \(f(U)\) and \(f(V)\) form a separation of \(Y \setminus f(S)\) and \(f(S)\) is a cut set of \(Y\text{.}\)
Activity 18.8.
(a)
Use the idea of cut sets/points to explain why the unit circle in \(\R^2\) is not homeomorphic to the interval \([0,1]\) in \(\R\text{.}\) Note: the unit circle is the set \(\{(x,y) \mid x^2+y^2 = 1\}\text{.}\) Draw pictures to illustrate your explanation. (A formal proof is not necessary, but you need to provide a convincing justification.)
(b)
Consider the following subsets of \(\R^2\) in the subspace topology:
\begin{equation*}
A = \{ (x,y) \mid x^2+y^2 =1\} \ \ \ \text{ and } \ \ \ B = \{ (x,0) \mid -1 \leq x \leq 1\}\text{.}
\end{equation*}
Is \(A \cup B\) homeomorphic to \(A\text{?}\) (A formal proof is not necessary, but you need to provide a convincing justification.)
We have seen that topological equivalence is an equivalence relation, which partitions the collection of all topological spaces into disjoint homeomorphism classes. Topological invariants can then help us identify the classes to which different spaces belong. In general, though, it can be more difficult to prove that two spaces are homeomorphic than not homeomorphic.
Activity 18.9.
Consider the spaces \(S_1 = \R\text{,}\) \(S_2 = (0,1)\) in \(\R\text{,}\) \(S_3 = [-1,1]\) in \(\R\text{,}\) the line segment \(S_4\) in \(\R^2\) between the points \((0,0)\) and \((2,2)\text{,}\) the space \(S_5\) determined by the letter { X}, and the space \(S_6\) determined by the letter { Y} in \(\R^2\text{.}\) Identify the distinct homeomorphism classes determined by these six spaces. No formal proofs are necessary, but you need to give convincing arguments.
