[Munkres Topology] 16. The Subspace Topology
This article is one of the posts in the Textbook Commentary Project.
Definition. Let $X$ be a topological space with topology $\mathcal{T}$. If $Y$ is a subset of $X$, the collection $$\mathcal{T}_Y = \{ Y \cap U | U \in \mathcal{T}\}$$ 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 $Y$.
It is easy to see that $\mathcal{T}_Y$ is a topology. It contains $\varnothing$ and $Y$ because $$\varnothing =Y \cap \varnothing \quad \mbox{and} \quad Y=Y\cap X,$$ 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 $$(U_1 \cap Y) \cap \cdots \cap (U_n\cap \cdots \cap U_n)\cap Y,$$ $$\bigcup_{\alpha \in J} (U_\alpha \cap Y) = (\bigcap_{\alpha \in J} U_\alpha) \cap Y.$$
Lemma 16.1. If $\mathcal{B}$ is a basis for the topology of $X$ then the collection $$\mathcal{B}_Y=\{ \mathcal{B}\cap Y | B \in \mathcal{B}\}$$ 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 $B$ on $\mathcal{B}$ 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 $Y$.
When dealing with a space $X$ and a subspace $Y$, one needs to careful when one uses the term 'open set'. Does one mean an element of the topology of $Y$ or an element of the topology of $X$? We make the following definition: 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 $X$.
This is a special situation in which every set open in $Y$ is also open in $X$:
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 $Y\cap V$.
Now let us explore the relation between the subspace topology and the order and product topologies. For product topologies, the result is what one might expect; for order topologies, it is not.
Theorem 16.3. 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 $$(U\times V)\cap (A\times B) = (U \cap A)\times (V\cap B).$$ 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.
Now 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 $X$.
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. 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 $$(a,+\infty)\cap Y = \{ x| x \in Y \mbox{ and } x > a\};$$ 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 is contained in the subspace topology.
To avoid ambiguity, let us agree that whenever $X$ is an ordered set in the order topology and $Y$ is a subset of $X$, we shall assume that $Y$ is given the subspace topology unless we specifically state otherwise. If $Y$ is convex in $X$, this is the same as the order topology on $Y$; otherwise it may not be.