Basis for a Subspace Topology
Let \(X\) be a topological space with basis \(B_X\), and let \(Y \subseteq X\). A basis for the subspace topology on \(Y\) can be obtained directly from the basis of \(X\). Specifically, the collection $$ B_Y=\{B\cap Y \mid B\in B_X\} $$ is a basis for the subspace topology on \(Y\).
This result is one of the most useful tools in topology because it allows us to construct a basis for a subspace without having to determine all of its open sets explicitly.
Example
Let us see how this theorem works in practice.
Consider the set of real numbers \( \mathbb{R} \) equipped with its standard topology. A basis for this topology is the collection of all open intervals:
$$ B_{\mathbb{R}}=\{(a,b)\mid a,b\in\mathbb{R},\ a< b\}. $$
Now consider the subset
$$ Y=[0,2]. $$
Our goal is to find a basis for the subspace topology on \(Y\).
According to the theorem, we simply intersect every basis element of \( \mathbb{R} \) with the set \(Y\):
$$ B_Y=\{(a,b)\cap[0,2]\mid (a,b)\in B_{\mathbb{R}}\}. $$
Why does this work? Every open set in \( \mathbb{R} \) can be written as a union of open intervals. Since the subspace topology is obtained by intersecting open sets of \( \mathbb{R} \) with \(Y\), it is enough to intersect the basis elements themselves.
Let us look at a few examples.
If we take the interval \((-1,1)\), then
$$ (-1,1)\cap[0,2]=[0,1). $$
Therefore, \([0,1)\) is a basis element of the subspace topology on \(Y\).
Next, consider the interval \((1,3)\):
$$ (1,3)\cap[0,2]=(1,2]. $$
Thus, \((1,2]\) is another basis element.
Finally, consider the interval \((0.5,1.5)\):
$$ (0.5,1.5)\cap[0,2]=(0.5,1.5). $$
In this case, the interval already lies entirely inside \(Y\), so the intersection leaves it unchanged.
These examples reveal an important feature of subspace topologies. Sets such as \([0,1)\) and \((1,2]\) are not open in \( \mathbb{R} \), but they are open in the subspace \(Y=[0,2]\).
Why These Sets Form a Basis
To understand why \(B_Y\) is a basis, let us examine an open set in the subspace topology.
Consider
$$ W=(0.5,1.5]. $$
This set is open in \(Y\) because it can be expressed as the intersection of \(Y\) with an open set of \( \mathbb{R} \):
$$ W=(0.5,2)\cap[0,2]. $$
Now choose any point of \(W\), for example \(y=1\).
Since \((0.5,2)\) is open in \( \mathbb{R} \), there exists a basis element of \( \mathbb{R} \) that contains \(y\) and is entirely contained in \((0.5,2)\). One possible choice is
$$ (0.8,1.2). $$
Intersecting this interval with \(Y\) gives
$$ (0.8,1.2)\cap[0,2]=(0.8,1.2). $$
This set belongs to \(B_Y\), contains the point \(y=1\), and lies completely inside \(W\).
Exactly the same argument works for every point of \(W\). Therefore, every point of an open set in the subspace topology is contained in a basis element that remains inside that open set. This is precisely the defining property of a basis.
Proof
Let \(X\) be a topological space with basis \(B_X\), and let \(Y\subseteq X\).
Define
$$ B_Y=\{B\cap Y\mid B\in B_X\}. $$
We will show that \(B_Y\) is a basis for the subspace topology on \(Y\).
First, every set \(B\cap Y\) is open in the subspace topology because it is the intersection of \(Y\) with an open set \(B\) of \(X\).
Now let \(W\) be an open set in \(Y\).
By the definition of the subspace topology, there exists an open set \(U\subseteq X\) such that
$$ W=U\cap Y. $$
Choose any point \(y\in W\).
Since \(y\in U\) and \(U\) is open in \(X\), the definition of a basis guarantees the existence of some basis element \(B\in B_X\) such that
$$ y\in B\subseteq U. $$
Intersecting with \(Y\), we obtain
$$ y\in B\cap Y\subseteq U\cap Y=W. $$
Since \(B\cap Y\in B_Y\), every point of \(W\) belongs to an element of \(B_Y\) that is entirely contained in \(W\).
Consequently, every open set in the subspace topology can be written as a union of elements of \(B_Y\).
Therefore, \(B_Y\) is a basis for the subspace topology on \(Y\).
