defines a topology on \(Y\) called the subspace topology. We call \(Y\text{,}\) together with the topology \(\mathcal{T}_Y\text{,}\) a subspace of \(X\text{.}\)
Proof.
We show that the collection \(\mathcal{T}_Y\) is a topology on \(Y\text{.}\)
Axiom (1). Since \(\emptyset, X\in \mathcal{T}\in X\text{,}\) we have \(\emptyset=\emptyset\cap Y\in \mathcal{T}_Y\) and \(Y=X\cap Y\in \mathcal{T}_Y\text{.}\)
Axiom (2). Let \(\{V_i\}_{i\in I}\) be a collection of elements of \(\mathcal{T}_Y\text{.}\) By definition we have \(V_i=U_i\cap Y\) for open sets \(U_i\in \mathcal{T}\text{.}\) It follows that
since \(U=\bigcup_{i\in I}U_i\) is open in \(X\text{.}\)
Axiom (3). Given elements \(V_1=U_1\cap Y, V_2=U_2\cap Y\in \mathcal{T}_Y\text{,}\) we have \(V_1\cap V_2=(U_1\cap U_2)\cap Y\in \mathcal{T}_Y\text{.}\)
Theorem1.6.1.Subspace properties.
Let \(Y\) be a subspace of \((X,\mathcal{T})\text{.}\)
If \(\mathcal{B}\) is a basis of \(\mathcal{T}\text{,}\) then the set \(\mathcal{B}_Y=\{B\cap Y\colon B\in\mathcal{B}\}\) is a basis for \(Y\text{.}\)
\(C\subseteq Y\) is closed in \(Y\) if and only if \(C=C'\cap Y\) for some \(C'\subseteq X\) that is closed in \(X\text{.}\)
If \(X\) is \(T_1\) (resp. Hausdorff, resp. metrizable), then \(Y\) is \(T_1\) (resp., Hausdorff, resp. metrizable).
Proof.
Axiom 1. For any \(y\in Y\) we have \(y\in B\) for some \(B\in \mathcal{B}\text{,}\) and thus \(y\in B\cap Y\in \mathcal{B}_Y\text{.}\)
Axiom 2. Given two elements \(B_1\cap Y, B_2\cap Y\in \mathcal{B}_Y\text{,}\) and \(y\in (B_1\cap Y)\cap (B_2\cap Y)=(B_1\cap B_2)\cap Y\text{,}\) there is a basis element \(B_3\in \mathcal{B}\) satisfying \(y\in B_3\subseteq B_1\cap B_2\text{.}\) It follows that \(B_3\cap Y\in \mathcal{B}_Y\) and \(y\in B_3\cap Y\subseteq (B_1\cap Y)\cap (B_2\cap Y)\text{.}\)
Let \(C\subseteq Y\text{.}\) We have
\begin{align*}
C \text{ closed in } Y \amp \iff Y-C \text{ open in } Y \\
\amp\iff Y-C=U\cap Y \text{ for some open } U\subseteq X\\
\amp \iff C=(X-U)\cap Y \text{ for some open } U\subseteq X\\
\amp \iff C=C'\cap Y \text{ for some closed } C'\subseteq X\text{.}
\end{align*}
Assume \(X\) is \(T_1\text{.}\) Given \(y\in Y\text{,}\) the set \(C'=\{y\}\) is closed in \(X\text{.}\) From the previous result it follows that \(\{y\}=\{y\}\cap Y\) is closed in \(Y\text{.}\) Thus \(Y\) is \(T_1\text{.}\)
Assume \(X\) is Hausdorff. Given \(y_1\ne y_2\in Y\) there are disjoint open (in \(X\)) neighborhoods \(U_1, U_2\) of \(y_1,y_2\text{,}\) respectively. The sets \(V_1=U_1\cap Y, V_2=U_2\cap Y\) are then open disjoint neighborhoods of \(y_1, y_2\) in the subspace topology of \(Y\text{.}\)
Example1.6.2.Open/closed depends on topology.
Let \(Y\) be a subspace of \(X\text{.}\) Prove:
\begin{align*}
U\text{ open in } Y \amp\not\!\!\!\!\implies U \text{ open in } X \\
C\text{ closed in } Y \amp\not\!\!\!\!\implies C \text{ closed in } X
\end{align*}
Solution.
Consider \(K=\{1/N\colon N\in \Z_+\}\) as a subspace of \(\R\) with the standard topology. The set \(K\) it self is both open and closed in the subspace \(K\text{,}\) but is neither open nor closed in \(\R\text{.}\)
Theorem1.6.3.Openness/closedness transitivity.
Let \(Y\) be a subspace of \(X\text{.}\)
If \(U\) is open in \(Y\) and \(Y\) is open in \(X\text{,}\) then \(U\) is open in \(X\text{.}\)
If \(C\) is closed in \(Y\) and \(Y\) is closed in \(X\text{,}\) then \(C\) is closed in \(X\)
Proof.
If \(U\) is open in \(Y\text{,}\) then \(U=Y\cap V\) for some open set \(V\) of \(X\text{.}\) Since \(Y\) is open in \(X\text{,}\) we have \(U=Y\cap V\) is open in \(X\text{.}\)
The proof is obtained from the previous argument by replacing all instances of “open” with “closed”, using (2) of Theorem 1.6.1.
SubsectionFinite product spaces
Topological specimen10.Finite products.
Let \(X_1, X_2,\dots, X_n\) be topological spaces, and let
\begin{equation*}
\mathcal{B}=\{U_1\times U_2\times \cdots \times U_n\colon U_i \text{ open in } X_i\}
\end{equation*}
is a basis on \(X\text{.}\) The topology \(\mathcal{T}\) generated by \(\mathcal{B}\) is the product topology on \(X\text{,}\) and \(X\) is called the product of the spaces \(X_1, X_2, \dots, X_n\text{.}\)
Proof.
We show that \(\mathcal{B}\) is a basis.
Axiom 1. Given \(x=(x_1,x_2,\dots, x_n)\in X\text{,}\) we have \(x\in X_1\times X_2\times\cdots\times X_n\in \mathcal{B}\text{.}\)
Let \(B=U_1\times U_2\cdots \times U_n\) and \(B'=V_1\times V_2\times \cdots \times V_n\text{.}\) Given \(x=(x_1,x_2,\dots, x_n)\in B\cap B'\text{,}\) we have \(x_i\in U_i\cap V_i\) for all \(1\leq i\leq n\text{,}\) and hence \(x\in B''=(U_1\cap V_1)\times (U_2\cap V_2)\times \cdots \times (U_n\cap V_n)\in \mathcal{B}\text{.}\)
Example1.6.4.Product basis not a topology necessarily.
Consider the product topology \(\mathcal{T}\) on \(X=\R^2=\R\times \R\text{,}\) where each copy of \(\R\) is equipped with the standard topology.
Describe the open sets of \(\mathcal{T}\) qualitatively. Give an example demonstrating that the product basis is not a topology in this case.
Show that \(\mathcal{T}\) is equal to the the Euclidean topology.
Solution.
Firstly, as we will show in more generality in Theorem 1.7.4, the set \(\mathcal{B}=\{(a,b)\times (c,d)\colon a < b, c < d\}\) forms a basis for the product topology on \(X=\R\times \R\text{.}\) Graphically, these basis elements are just open rectangles in the plane.
An open set in the product topology on \(\R^2\) is anything that can be written as a union of open rectangles, by the remark made above. To see that the product basis is not in general a topology, consider \(U_1\cup U_2\text{,}\) where \(U_1=(0,1)\times (0,1)\) and \(U_2=(2,3)\times (2,3)\text{.}\) We claim that not only is \(U_1\cup U_2\) not a product of open sets, it is in fact not the product of any sets. Indeed, if \(U_1\cup U_2=A\times B\) for two sets \(A,B\subseteq \R\text{,}\) then we would have to have \((0,1)\cup (2,3)\subseteq A, B\text{,}\) in which case \((0,1)\times (2,3)\subseteq A\times B\text{.}\) This is a contradiction since \((0,1)\times (2,3)\not\subseteq U_1\cup U_2\text{.}\)
Using the remark above, it is enough to show that open circles (basis elements in the standard topology) can be written as unions of open rectangles (basis elements in the product topology) and vice versa. This is clear.
