§ 16. The Subspace Topology¶
Definition 16.1 : Subspace Topology / Subspace
Let \(X\) be a topological space with topology \(\mathcal{T}\). If \(Y\) is a subset of \(X\), the collection
is a topology on \(Y\), called the subspace topology. With this topology, \(Y\) is called a subspace of \(X\); its open sets consist of all intersections of open sets of \(X\) with \(Y\).
Theorem 16.0
\(\mathcal{T}_{Y}\) is indeed a topology.
Proof
\(\mathcal{T}_{Y}\) contains \(\varnothing\) and \(Y\) because
where \(\varnothing\) and \(X\) are elements of \(\mathcal{T}\). The fact that it is closed under finite intersections and arbitrary unions follows from the equations
Lemma 16.1
If \(\mathcal{B}\) is a basis for the topology of \(X\) then the collection
is a basis for the subspace topology on \(Y\).
Proof
Given \(U\) open in \(X\) and given \(y \in U \cap Y\), we can choose an element \(B\) of \(\mathcal{B}\) such that \(y \in B \subset U\). Then \(y \in B \cap Y \subset U \cap Y\). It follows from Lemma 13.2 that \(\mathcal{B}_{Y}\) is a basis for the subspace topology on \(Y\).
Definition 16.2 : Open In
If \(Y\) is a subspace of \(X\), we say that a set \(U\) is open in \(Y\) (or open relative to \(Y\) ) if it belongs to the topology of \(Y\); this implies in particular that it is a subset of \(Y\). We say that \(U\) is open in \(X\) if it belongs to the topology of \(X\).
Lemma 16.2
Let \(Y\) be a subspace of \(X\). If \(U\) is open in \(Y\) and \(Y\) is open in \(X\), then \(U\) is open in \(X\).
Proof
Since \(U\) is open in \(Y, U=Y \cap V\) for some set \(V\) open in \(X\). Since \(Y\) and \(V\) are both open in \(X\), so is \(Y \cap V\).
Theorem 16.3
(Subspace topology \(\Leftrightarrow\) Product topology)
If \(A\) is a subspace of \(X\) and \(B\) is a subspace of \(Y\), then the product topology on \(A \times B\) is the same as the topology \(A \times B\) inherits as a subspace of \(X \times Y\).
Proof
The set \(U \times V\) is the general basis element for \(X \times Y\), where \(U\) is open in \(X\) and \(V\) is open in \(Y\). Therefore, \((U \times V) \cap(A \times B)\) is the general basis element for the subspace topology on \(A \times B\). Now
Since \(U \cap A\) and \(V \cap B\) are the general open sets for the subspace topologies on \(A\) and \(B\), respectively, the set \((U \cap A) \times(V \cap B)\) is the general basis element for the product topology on \(A \times B\).
The conclusion we draw is that the bases for the subspace topology on \(A \times B\) and for the product topology on \(A \times B\) are the same. Hence the topologies are the same.
Theorem 16.3.1
(Subspace topology \(\nLeftrightarrow\) Order topology)
Let \(X\) be an ordered set in the order topology, and let \(Y\) be a subset of \(X\).
The order relation on \(X\), when restricted to \(Y\), makes \(Y\) into an ordered set.
However, the resulting order topology on \(Y\) need not be the same as the topology that \(Y\) inherits as a subspace of \(X\).
Example
Let \(Y\) be the subset \([0,1) \cup\{2\}\) of \(\mathbb{R}\). In the subspace topology on \(Y\) the one-point set \(\{2\}\) is open, because it is the intersection of the open set \(\left(\frac{3}{2}, \frac{5}{2}\right)\) with \(Y\). But in the order topology on \(Y\), the set \(\{2\}\) is not open. Any basis element for the order topology on \(Y\) that contains 2 is of the form
for some \(a \in Y\); such a set necessarily contains points of \(Y\) less than 2.
Example
Let \(I=[0,1]\). The dictionary order on \(I \times I\) is just the restriction to \(I \times I\) of the dictionary order on the plane \(\mathbb{R} \times \mathbb{R}\). However, the dictionary order topology on \(I \times I\) is not the same as the subspace topology on \(I \times I\) obtained from the dictionary order topology on \(\mathbb{R} \times \mathbb{R}\) ! For example, the set \(\{1 / 2\} \times(1 / 2,1]\) is open in \(I \times I\) in the subspace topology, but not in the order topology, as you can check.
Definition 16.3 : Ordered Square
Let \(I=[0,1]\). The dictionary order on \(I \times I\) is just the restriction to \(I \times I\) of the dictionary order on the plane \(\mathbb{R} \times \mathbb{R}\). The set \(I \times I\) in the dictionary order topology will be called the ordered square, and denoted by \(I_{o}^{2}\).
Definition 16.4 : Convex
Given an ordered set \(X\), let us say that a subset \(Y\) of \(X\) is convex in \(X\) if for each pair of points \(a<b\) of \(Y\), the entire interval \((a, b)\) of points of \(X\) lies in \(Y\). Note that intervals and rays in \(X\) are convex in \(X\).
Theorem 16.4
(Subspace topology \(\Leftrightarrow\) Order topology) (In convex subspace)
Let \(X\) be an ordered set in the order topology; let \(Y\) be a subset of \(X\) that is convex in \(X\).
Then the order topology on \(Y\) is the same as the topology \(Y\) inherits as a subspace of \(X\).
Proof
Consider the ray \((a,+\infty)\) in \(X\). What is its intersection with \(Y\) ? If \(a \in Y\), then
this is an open ray of the ordered set \(Y\). If \(a \notin Y\), then \(a\) is either a lower bound on \(Y\)
or an upper bound on \(Y\), since \(Y\) is convex. In the former case, the set \((a,+\infty) \cap Y\) equals all of \(Y\); in the latter case, it is empty.
A similar remark shows that the intersection of the ray \((-\infty, a)\) with \(Y\) is either an open ray of \(Y\), or \(Y\) itself, or empty. Since the sets \((a,+\infty) \cap Y\) and \((-\infty, a) \cap Y\) form a subbasis for the subspace topology on \(Y\), and since each is open in the order topology, the order topology contains the subspace topology.
To prove the reverse, note that any open ray of \(Y\) equals the intersection of an open ray of \(X\) with \(Y\), so it is open in the subspace topology on \(Y\). Since the open rays of \(Y\) are a subbasis for the order topology on \(Y\), this topology is contained in the subspace topology.