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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 $\tau_A$.

These two methods should produce the same result as illustrated by the diagram below: $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} \gamma_X & \ra{open} & \tau_X \\ \da{\cap A} & & \da{\cap A} \\ \gamma_A & \ra{open} & \tau_A \end{array}$$

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. $\blacksquare$

A less direct but more elegant proof is to observe that the restriction of a map to a subset is its composition with the inclusion of the subset: $$f|_A=f i_A,$$ so that the diagram commutes: $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & A & \ra{i_A} & X \\ & _{f|_A} & \searrow & \da{f} \\ & & & Y \end{array}$$ Both are continuous and so is their composition.

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 $F|_A=f$.

Extension Problem. Is there a continuous $F$ to complete the diagram below so that it is commutative? $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} & A & \ra{i_A} & X \\ & _{f} & \searrow & \da{F=?} \\ & & & Y \end{array}$$ Unlike the restriction problem above, the extension problem can't be solved by simply forming the composition. In fact, such an extension might simply not exist.

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:

$$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccc} &\{0,1\} & \ra{i_A} & [0,1] & & H_0(\{0,1\}) & \ra{} & H_0([0,1]) & & < 0,1 > & \ra{} & < 0 >\\ & & _{f}\searrow & \da{F=?} & & _{} & \searrow & \da{?} & & & \searrow & \da{?}\\ & & & X & & & & H_0(X) & & & & H_0(X) \end{array}$$ These groups are evaluated in the last diagram. But the way we construct the homology groups and homology maps make us realize that we can complete the first diagram only if we can complete the last one. But to do that -- for any given $f$ -- $H_0(X)$ would have to be $1$-dimensional!

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...