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# Relative topology

## 1 How subsets inherit the topology

What is a relation between the topology of the $xy$-plane and the topology of the $x$-axis?

The question can be that of convergence: if $x_n\to a$ then $(x_n,0) \to (a,0)$:

Or that of continuity: if $f(\cdot,\cdot)$ is continuous then $f(\cdot,0)$ is continuous too:

Or it can be about interior, exterior, closure, etc.

There must be a relation!

Suppose $X$ be the $xy$-plane and $A$ is the $x$-axis. We know that the closeness/proximity of points in $X$ is measured simply by the distance formula, the Euclidean metric: $$d_X((x,y),(a,b))=\sqrt{(x-a)^2+(y-v)^2}.$$ But if these two points happen to lie on the $x$-axis, the formula turns into: $$d_A(x,a)=|x-a|,$$ which is the Euclidean metric of the $x$-axis. Therefore, the closeness/proximity of points in $A$ are measured in a way that matches that of $X$.

So, with metric spaces the approach is clear. 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 $a$.
• 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 $a$.

But what about non-metric topological spaces?

Back to our example.

• The Euclidean basis of $A$ is the open intervals.
• The Euclidean basis of $X$ is the open disks.

But intersections of disks with the $x$-axis produce intervals and, vice versa, all intervals in the $x$-axis are intersections of disks with it:

Of course, there are many different disks for each interval.

This idea applies to all topological spaces. Indeed, with our approach to topology via neighborhoods, a subset of a topological space with basis $\gamma$ will acquire its own collection as its intersections with the elements of $\gamma$:

## 2 How neighborhoods are inherited by a subset

Suppose $\gamma_X$ is a basis of neighborhoods in $X$. Then we define a collection of subsets of $A$, as above: $$\gamma_A = \{W \cap A \colon W \in \gamma_X \}.$$

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

Proof. Given $x \in A \in X$, then $x \in X$. Since $\gamma_X$ is a basis on $X$, it satisfies (B1): there is some $W \in \gamma_X$ 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 $\gamma_X$,

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). $\blacksquare$

The topology generated by $\gamma_A$ is called the relative topology on $A$ generated by the topology of $X$.

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

Suppose $X$ is Euclidean with nice round neighborhoods:

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

In fact, since it doesn't matter how $A$ used to fit into $X$ anymore, we can even rearrange its pieces:

## 3 The topology of a subpace

What if the topology is given by a collection $\tau$ of open sets?

In our example, all open sets in the $x$-axis $A$ can be seen as intersections of open sets in the $xy$-plane $X$:

This statement is less obvious and will need a proof.

Generally, 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, a collection of open sets $W$. Then,

what should be the topology for $A$?

All we need to do is to point out the set we deem open within the subset. We choose, every time, $W_A = W \cap A$ for all $W$ open in $X$.

Suppose $\tau_X$ 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 $\tau_X$.

For (T1):

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

For (T3):

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

since

• $W'=U' \cap V' \in \tau_X.$

(T2) Exercise. $\blacksquare$

In this case $A$ is a new topological space called a subspace of $X$.

Compare this idea to that of a subspace of a vector space. In either case, the structure (topological or algebraic) is inherited by a subset. The main difference is that for vector spaces (or groups) the subset needs to be closed under the operations while for topological spaces a subset can always be seen as a subspace.

Note. In illustrations above we make it point to choose a subset $A$ to be depicted as something "thin" in comparison to a "thick" $X$. The reason is that a thick $A$ might implicitly suggest that $A$ is an open subset of $X$ and that might make you reply on this subconscious assumption in your proofs.

Exercise. What is so special about the relation between $\tau_X$ and $\tau_A$ when $A$ is an open subset of $X$?

Once the topology on $A$ is established, all the topological concepts become available. We speak of open and closed sets "in $A$" (as opposed to "in $X$"), interior, exterior, closure, frontier, continuity, etc.

The following is obvious.

Theorem. The sets closed in $A$ are the intersections of the closed sets in $X$ with $A$.

Exercise. If $A$ is closed in $X$ then sets closed in $A$ are...

Theorem. If $x_n\to a$ in $A\subset X$ then $x_n \to a$ in $X$.

Exercise. Prove the theorem.

The converse isn't true. While $x_n=1/n$ converges to $0$ in $X={\bf R}$, it diverges in $A=(0,1)$. This fact affects the limits points of subsets and, therefore, the closure. Here's a simple example that illustrates the difference between the closure in $X$ and the closure in $A \subset X$. Suppose

• $X = {\bf R}, A = (0,1), B = (0,1),$

then

• in $X$, ${\rm Cl}(B) = [0,1]$,
• in $A$, ${\rm Cl}(B) = (0,1).$

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

Exercise. Describe the relative topology of the following subsets of $X={\bf R}$:

1. $A=\{1,2,...,n\}$,
2. $A={\bf Z}$,
3. $A=\{1,1/2,1/3,...\}$,
4. $A=\{1,1/2,1/3,...\}\cup \{0\}$,
5. $A={\bf Q}$.

Exercise. Show that if $Q$ is a subspace of $A$ and $A$ is a subspace of $X$, then $Q$ is a subspace of $X$.

Exercise. (a) Suppose $A,B$ are disjoint subsets of topological space $X$. Compare these three topologies:

• $A$ relative to $X$,
• $B$ relative to $X$, and
• $A\cup B$ relative to $X$.

(b) Suppose two (unrelated to each other) topological spaces $X$ and $Y$ are given. Based on part (a), discuss what the definition of the topology of the disjoint union $X \sqcup Y$ of these spaces ought to be? What happens to the bases?

## 4 Relative neighborhoods vs relative topology

An important but easy to miss question remains, do $\gamma_A$ and $\tau_A$ match?

What do we mean by "match"? Starting with $\gamma_X$, we have acquired $\tau_A$ in two different ways:

1. taking all sets open with respect to $\gamma_X$ creates $\tau_X$, then taking the latter's intersections with $A$ creates $\tau_A$; or
2. taking intersections with $A$ of $\gamma_X$ creates $\gamma_A$, then taking all sets open with respect to the latter creates $\tau_A$.


In particular, the question from the beginning of this section needs to be answered: why are all open sets in the $x$-axis are the intersections of all open sets in the $xy$-plane with the $x$-axis?

Theorem.

• $\tau_A$
• = the set of all intersections of the elements of $\tau_X$ with $A$
• = the set of all sets open with respect to $\gamma_A$.

Proof. To prove the equality we prove the mutual inclusions for these two sets.

Part 1: $\subset$. Given $U\in \tau_A$, show that $U$ is open with respect to $\gamma_A$. Suppose $x\in U$, we need to find $N\in \gamma_A$ such that $x\in N\subset U$. Of course we construct as the intersection with a neighborhood of $x$ in $X$. Since $U$ is open in $A$, there is an open $U'$ in $X$ such that $U=U'\cap A$. Since $U'$ is open there is a neighborhood $N\in \gamma_X$ such that $x\in N'\subset U'$. Now we take $N=N'\cap A$.

Part 2: $\supset$. It is less straight-forward and will require a subtle step. Suppose $U\in \tau_A$, i.e., it's open with respect to $\gamma_A$. We need to find $U'$ open in $X$ such that $U=U'\cap A$. Since $U$ is open in $A$, for every $x\in U$ there is $N_x\in\gamma_A$ such that $x\in N_x\in U$. Further, $N_x=N'_x\cap A$ for some $N'_x\in \gamma_X$. But even though this set is open in $X$, it can't possibly give us the desired $U'$. The ingenious idea is to take all of these neighborhoods together: $$U'= \bigcup _{x\in U} N'_x.$$ It is open by (T2).

$\blacksquare$

## 5 New maps

In light of this new concept of relative topology we can restate a recent theorem.

Theorem. Suppose $f \colon X \rightarrow Y$ is continuous. If $X$ is path-connected the so is $f(X)$. In other words, the continuous image of a path-connected space is path-connected.

As a generalization of piecewise defined functions, two continuous functions can be "glued together" to create a new continuous function, as follows.

Theorem (Pasting lemma). Let $A,B$ be two closed subsets of a topological space $X$ such that $X = A \cup B$, and let $Y$ also be a topological space. Suppose $f_A: A \to Y,f_B:B\to Y$ are continuous functions, and $$f_A(x)=f_B(x),\forall x\in A\cap B.$$ Then the function $f:X\to Y$ defined by $$f(x)= \begin{cases}{} f_A(x) & \text{ if } x\in A, \\ f_B(x) & \text{ if } x\in B, \end{cases}$$ is continuous.

Observe that the theorem fails if $A,B$ aren't closed. Just consider the example of the sign function.

Exercise. Prove the lemma. Hint: you'll have to use relative topology. State and prove the analogue of the theorem with "closed" replaced by "open".

Exercise. Prove that if $f:X\to Y$ is continuous then so is $f':X\to f(X)$ given by $f'(x)=f(x),\forall x \in X$.

Definition. A one-to-one map $f:X\to Y$ the partial inverse $f^{-1}:f(X) \to X$ of which is also continuous is called an embedding.

Exercise. Prove that an embedding $f:X\to Y$ creates a homeomorphism $f':X\to f(X)$.

Suppose we have a topological space $X$ and a subset $A$ of $X$. Then the inclusion function $i=i_A: A {\rightarrow} X$ of $f$ is given by $$i_A(x) = x,\forall x \in A.$$

For an open set $U$ in $X$ $$i_A^{-1}(U)=U\cap A,$$ so from the definition of relative topology, it follows:

Theorem. The inclusion function is continuous.

The notation often used for the inclusion is $$i=i_A:A \hookrightarrow X.$$

Next, notice that the picture below suggests that limiting the domain of a function will preserve its continuity:

Suppose we have topological spaces $X$ and $Y$, a map $f: X \rightarrow Y$, and a subset $A$ of $X$. Then the restriction $f|_A$ of $f$ to $A$ is a function $$f|_A: A {\rightarrow} Y$$ defined by $$f|_A(x) = f(x), \forall x \in A.$$

In particular, we can understand inclusions as restrictions of the identity functions.

Theorem. A restriction of a continuous function is continuous.

Proof. Suppose we have a continuous function $f \colon X \rightarrow Y$ and a subset $A$ of $X$. Suppose $U$ is open in $Y$. Then $$(f|_A)^{-1}(U) = f^{-1}(U) \cap A.$$ As the intersection of an open, in $X$, set with $A$, this set is open in $A$. Hence $f|_A$ is continuous. $\blacksquare$


Exercise. Suppose $i_A:A\to X$ is he inclusion. Suppose the set $A$ is given a topology such that for every topological space $Y$ and very function $f:Y\to A$,

$f: Y\to A$ is continuous $\Leftrightarrow$ the composition $i_Af:Y\to X$ is continuous.

Prove that this topology coincides with the relative topology of $A$ in $X$.

## 6 The extension problem

Just as we can restrict maps, we can try to "extend" them, continuously!

For $A\subset X$ and a given function $f:A\to Y$, a function $F:X\to Y$ is called an extension of $f$ if $F|_A=f$.


Of course, it is simple to extend a continuous function $f:[0,1]\to {\bf R}$ to $F:[0,2]\to {\bf R}$. But what about $f:(0,1)\to {\bf R}$ and $F:[0,1]\to {\bf R}$? For $f(x)=x^2$, yes; for $f(x)=1/x$ or $f(x)=\sin (1/x)$, no.

A more interesting example comes from our recent study. We can recast the definition of path-connectedness of a topological space $X$ as the solvability of a certain extension problem: can every map defined on the end-points of the interval $$f:\{0,1\} \to X$$ be extended to a map on the whole interval $$F:[0,1] \to X?$$

Let's sneak a peak into how this approach will be used in algebraic topology. The first two diagrams below are:

• the extension diagram -- with topological spaces and maps, and
• its algebraic counterpart -- with groups and homomorphisms:


The worth of this approach becomes clearer as we start climbing dimensions. For example, we can also recast the definition of simple connectedness of a topological space $X$ as an extension problem: can every map of the circle $$f:{\bf S}^1 \to X$$ be extended to a map of the disk $$F:{\bf B}^2 \to X?$$

And so on for all dimensions, even those impossible to visualize...