K. Jänich, Topology. Chapter I: Fundamental Concepts. 3: Subspaces, Disjoint Unions and Products

Definition 1: Subspace Let \((X, \mathcal{O})\) be a topological space and \(Y \subseteq X\). The subspace topology on \(Y\) is \(\{ U \cap Y : U \in \mathcal{O} \}\). A space of the form \((Y,\mathcal{O}')\) where \(Y \subseteq X\) and \(\mathcal{O}'\) is the subspace topology is called a subspace of \(X\).

If \(X\) is a topological space, \(Y\) is a subset of \(X\), and we talk about \(Y\) as a topological space, it is to be assumed that it has the subspace topology. We say a set is open in \(Y\) iff it is open in the subspace topology on \(Y\).

We prove that the subspace topology is a topology.

Proof Let \(\mathcal{O}' = \{ U \cap Y : U \in \mathcal{O} \} \).

  1. Let \( \mathcal{U} \subseteq \mathcal{O}' \). Then \( \bigcup \mathcal{U} = \bigcup \{ U \in \mathcal{O} : U \cap Y \in \mathcal{U} \} \cap Y \) and so \( \bigcup \mathcal{U} \in \mathcal{O}' \).
  2. Let \(U, V \in \mathcal{O}' \). Pick \(U',V' \in \mathcal{O}\) such that \(U = U' \cap Y\) and \(V = V' \cap Y\). Then \(U \cap V = U' \cap V' \cap Y\) and so \(U \cap V \in \mathcal{O}'\) since \(U' \cap V' \in \mathcal{O} \).
  3. We have \( Y = X \cap Y \in \mathcal{O}'\) since \(X \in \mathcal{O} \). \(\Box\)

Definition 2 Let \((X,\mathcal{O})\) and \((Y, \mathcal{O}')\) be topological spaces. We give the disjoint union \(X + Y\) the topology \( \{ U + V : U \in \mathcal{O}, V \in \mathcal{O}' \} \) and call \( X + Y \) the topological disjoint union of \(X\) and \(Y\) under this topology.

We prove this is a topology.

Proof Let \( \mathcal{O}_+ = \{ U + V : U \in \mathcal{O}, V \in \mathcal{O}' \} \).

  1. Let \( \mathcal{U} \subseteq \mathcal{O}_+ \). Then \( \bigcup \mathcal{U} = (\bigcup \{ U \in \mathcal{O} : \exists V \in \mathcal{O}'. U + V \in \mathcal{U} \}) + (\bigcup \{ V \in \mathcal{O}' : \exists U \in \mathcal{O}. U + V \in \mathcal{U} \})\) and so \( \bigcup \mathcal{U} \in \mathcal{O}_+ \).
  2. Given \(U_1 + V_1, U_2 + V_2 \in \mathcal{O}_+\) we have \( (U_1 + V_1) \cap (U_2 + V_2) = (U_1 \cap U_2) + (V_1 \cap V_2) \in \mathcal{O}_+ \).
  3. We have \( X + Y \in \mathcal{O}_+ \) because \( X \in \mathcal{O} \) and \(Y \in \mathcal{O}'\). \(\Box\)

Definition 3 Let \(X\) and \(Y\) be topological spaces. The product topology on \(X \times Y\) is defined by: a set \(W \subseteq X \times Y\) is open iff, for all \((x,y) \in W\), there exist neighbourhoods \(M\) of \(x\) in \(X\) and \(N\) of \(y\) in \(Y\) such that \(M \times N \subseteq W\).

We prove that this defines a topology.

Proof

  1. Let \(\mathcal{U}\) be a set of open sets in \(X \times Y\); we prove \(\bigcup \mathcal{U}\) is open. Let \((x,y) \in \bigcup \mathcal{U}\). Pick \(U \in \mathcal{U}\) such that \((x,y) \in U\). Pick neighbourhoods \(M\) of \(x\) and \(N\) of \(y) such that \(M \times N \subseteq U\). Then \(M \times N \subseteq \bigcup \mathcal{U}\) as required.
  2. Let \(U\) and \(V\) be open sets in \(X \times Y\); we prove \(U \cap V\) is open. Let \((x,y) \in U \cap V\). Pick neighbourhoods \(M_1\) of \(x\) and \(N_1\) of \(y\) such that \(M_1 \times N_1 \subseteq U\). Pick neighbourhoods \(M_2\) of \(x\) and \(N_2\) of \(y\) such that \(M_2 \times N_2 \subseteq V\). Then \(M_1 \cap M_2\) is a neighbourhood of \(x\) and \(N_1 \cap N_2\) is a neighbourhood of \(y\), and we have \((M_1 \cap M_2) \times (N_1 \cap N_2) \subseteq U \cap V \) as required.
  3. To prove \(X \times Y\) is open, let \((x,y) \in X \times Y\). Then \(X\) is a neighbourhood of \(x\) and \(Y\) is a neighbourhood of \(y\), and \(X \times Y \subseteq X \times Y\) as required. \(\Box\)

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