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# Relative topology

Suppose we have a set $X$ and suppose $A$ is a subset of $X$. As we know there are many ways to define a topology on $A$, if it is just another set. But suppose $X$ already has a topology (i.e., it is a topological space). Then,

what should be the topology for $A$?

Let's start with Euclidean space. Suppose $X = {\bf R}^n$ and $A$ is its subset. The closeness/proximity of points in $X$ is measured simply by the distance. Therefore, the closeness/proximity of points in $A$ should be measured the same way, in order to create a topology on $A$ that "matches" that of $X$.

Suppose we are given a point $a$ in $A$.

If we treat a as a point in $X$, then

a neighborhood $W$ of $a$ in $X$ consists of points in $X$ within some $d$ from $a$.

Now, if we treat $a$ as a point in $A$, then

a neighborhood $W_A$ of $a$ in $A$ consists of points in $A$ within some $d$ from $a$.

Then, $W_A = W \cap A$.

Now suppose $\gamma$ is a basis in $X$, then we define a collection of subsets of $A$ as above: $$\gamma_A = \{W \cap A \colon W \in \gamma \}.$$

Theorem. $\gamma_A$ is a basis in $A$.

Proof. Given $x \in A$, then $x \in X$. Since $\gamma$ is a basis of $X$, it satisfies (B1): there is some $W \in \gamma$ such that $x \in W$. Thus, $x \in A$ and $x \in W$, hence $x \in A \cap W \in \gamma_A$. Therefore, $\gamma_A$ satisfies (B1).

We need to prove (B2). Suppose we are given $U,W \in \gamma_A$ and a point $x \in U \cap W$. We need to prove that there is some $V \in \gamma_A$ such that $x \in V \subset U ∩ W$.

By definition of $\gamma_A$,

$U = U' \cap A, W = W' \cap A$ for some $U',W' \in \gamma.$

By (B2) for $\gamma$,

there is some $V' \in \gamma$ such that $x \in V' \subset U' \cap W'$.

Now let $$V = V' \cap A.$$ Thus, $x \in A$ and $x \in V$, hence $$x \in A \cap V = V' \in \gamma_A.$$ Also, $V' \subset U' \cap W'$ implies that $$V = V' \cap A' \subset U' \cap W' \cap A = (U' \cap A) \cap (W' \cap A) = U \cap W.$$ Therefore, $\gamma_A$ satisfies (B2). $\blacksquare$

Next the topology of $A$ can be generated by the basis. Alternatively, we can define the topology directly, in a way identical to what we did for the basis.

Suppose $\tau$ is the topology of $X$, then we define a collection of subsets of $A$: $$\tau_A = \{W \cap A \colon W \in \tau \}.$$

Theorem. $\tau_A$ is a topology in $A$.

Proof. (T1)-(T3) for $\tau_A$ follow from (T1)-(T3) for $\tau$.

For (T1):

• $\emptyset \in \tau \Rightarrow \emptyset = A \cap \emptyset \in \tau_A,$
• $X \in \tau \Rightarrow A = A \cap X \in \tau_A.$

For (T2):

• $U, V \in \tau_A \Rightarrow$
• $U = U' \cap A, W = W' \cap A$ for some $U', W' \in \gamma \Rightarrow$
• $U \cap W = (U' \cap A) \cap (W' \cap A) = (U' \cap W') \cap A \in \tau_A,$

since

• $U' \cap W' \in \tau.$

(T3) Exercise. $\blacksquare$

Let's review what happens to the neighborhoods in $X$ as we pass them over to $A$.

$X$ is Euclidean with nice round neighborhoods:

Now the neighborhoods in $A$ are explicit and $X$ is gone:

In fact, it does no matter how $A$ used to fit into $X$ anymore, so we can rearrange its pieces:

Once the topology on is established, all the topological concept become available: closed sets, interior, exterior, closure, frontier, continuity, etc.

Here's a simple example that illustrate the difference between closure in $X$ and the closure in $A \subset X$. $$X = {\bf R}, A = (0,1), B = (0,1), {\rm \hspace{3pt} then \hspace{3pt} in \hspace{3pt}} X, {\rm Cl}(B) = [0,1], {\rm \hspace{3pt} in \hspace{3pt}} A, {\rm Cl}(B) = (0,1).$$

Exercise. Find a non-euclidean topology (or a basis) on $X = {\bf R}^2$ that generates the euclidean topology on $A =$ the $x$-axis. Solution

The inclusion function $i_A \colon A \rightarrow X$ given by $i_A(x) = x$ is continuous. For more see Examples of maps.