Basic point set topology

Show that the following reformulation of continuity, Definition 1.1.2, is equivalent to the original definition, Definition 1.1.1. This requires showing that, if $f$ is continuous in the sense of Definition 1.1.1, then it is also continuous in the sense of Definition 1.1.2, and vice versa.

Definition 1.1.1: $f$ is continuous at $x \in X$ if, given $\epsilon > 0$, there is a $\delta > 0$ so that $d(x, y) < \delta$ implies that $d(f(x), f(y)) < \epsilon$. Here $d$ indicates the Euclidean distance function $d((x_1, \ldots , x_k), (y_1, \ldots , y_k)) = ((x_1 - y_1)^2 + \cdots + (x_k - y_k)^2)^{1/2}$. We say that $f$ is continuous if it is continuous at $x$ for all $x \in X$.

For $z \in \mathbb {R}^{k}$, define the ball of radius $r$ about $z$ to be the set $B(z, r) = \left\{ y \in \mathbb {R}^{k} : d(z, y) < r\right\}$. If $C$ is a subset of $\mathbb {R}^{k}$ and $z \in C$, denote $B_C(x, r) = C \cap B(x, r) = \left\{ y \in C : d(y, x) < r\right\}$.

Definition 1.1.2: $f : X \to Y$ is continuous at $x \in X$ if given $\epsilon > 0$, there is a $\delta > 0$ so that $B_X(x, \delta ) \subset f^{-1}(B_Y(f(x), \epsilon ))$. $f$ is continuous if it is continuous at $x$ for all $x \in X$.

Determine whether the following subsets of the plane are open. Justify your answers.

Verify that the definition of an open set for a metric space satisfies the requirements for a topology.

Let $d$ denote the usual Euclidean metric in the plane and let $d'((x, y), (u, v)) = \left|x - u\right| + \left|y - v\right|$. Verify that $d'$ is a metric.

The points in a space are continuous functions defined on the interval $[0, 1]$. Define the distance between two such functions to be

Show that this definition of the distance between two functions does satisfy the three properties required of a metric. This depends on the fact, which you may assume in your argument, that the integral of a nonnegative continuous function is positive unless the function is identically $0$.

For $X \subset \mathbb {R}^{n}$, show that the usual topology on $X$ is the same as the subspace topology.

Show that $[a, b]$ is a closed set in $\mathbb {R}$. Show that a rectangle (including the perimeter) is a closed set in $\mathbb {R}^{2}$.

Show that $[a, b)$ is neither open nor closed in $\mathbb {R}$.

Proposition 1.2.1: $f : A \to B$ is continuous iff the inverse images of closed sets are closed.

One direction has already been shown: suppose $f$ is continuous and $C$ is a closed subset of $B$. Then $B \setminus C$ is open and $f^{-1}(C) = A \setminus f^{-1}(B \setminus C)$ is closed since it is the complement of an open set in $A$.

Complete the proof by proving the converse: suppose that the inverse image of every closed set in $B$ is closed in $A$, and show that $f$ is continuous.

Proposition 1.2.2: If $A \subset X$ has the subspace topology, then $D \subset A$ is closed in $A$ iff $D = A \cap E$, where $E$ is closed in $X$.

One direction has already been shown: by the definition of the subspace topology, the open sets in $A$ are the intersections of $A$ with the open sets in $X$. Suppose $D$ is closed in $A$. Then $D = A \setminus F$, where $F$ is open in $A$. Then $F = A \cap G$, where $G$ is open in $X$. Hence, if $E = X \setminus G$, then $E$ is closed in $X$ and

Complete the proof by proving the converse: suppose $D = A \cap E$ where $E$ is closed in $X$, and show that $D$ is closed in $A$.

Suppose $A$ is a closed subset of $X$. Then $D \subset A$ is closed in $A$ (with the subspace topology) iff $D$ is closed in $X$.

The closure of a set $A \subset X$, denoted $\overline{A}$, is the intersection of all closed sets containing $A$. The interior of $A$, denoted ${A}^{\mathrm{o}}$, is the union of all open sets contained in $A$.

Show that $\overline{A}$ is closed and ${A}^{\mathrm{o}}$ is open.

Find $\overline{A}$, ${A}^{\mathrm{o}}$, and $\operatorname {Bd} A$ for the following sets $A$ in $\mathbb {R}^{2}$, where $\operatorname {Bd} A = \overline{A} \cap \overline{X \setminus A}$ denotes the boundary of $A$:

Show that $\overline{A} = {A}^{\mathrm{o}} \cup \operatorname {Bd} A$ and ${A}^{\mathrm{o}} \cap \operatorname {Bd} A = \emptyset$, where $\operatorname {Bd} A = \overline{A} \cap \overline{X \setminus A}$ denotes the boundary of $A$.

The piecing lemma (Lemma 1.2.3) states: suppose $X = A \cup B$, where $A$ and $B$ are closed subsets of $X$. Let $f : X \to Y$ be a function so that the restrictions of $f$ to $A$ and $B$ (given the subspace topology) are each continuous. Then $f$ is continuous.

Prove the analog of this lemma where the word closed'' is replaced by the word open.'' Give an example to show that the conclusion that $f$ is continuous is not true without some hypothesis on the sets $A, B$.

The piecing lemma for homeomorphisms states: suppose that $X = A \cup B$, $Y = C \cup D$, where $A, B$ are closed in $X$, and $C, D$ are closed in $Y$. Let $f : A \to C$ and $g : B \to D$ be homeomorphisms, and suppose that the restrictions of $f$ and $g$ to the intersection $A \cap B$ agree as maps into $Y$. Define $h : X \to Y$ by $h|_A = f$ and $h|_B = g$ (or we could start with $h$ and define $f$ and $g$ just by restricting them to $A$ and $B$). If $h$ is a bijection (this just requires that the only points that are in the image of both $f$ and $g$ are the points in the image of $A \cap B$), then $h$ is a homeomorphism.

Prove the piecing lemma for homeomorphisms.

A rigid motion (isometry) of the plane is a map $f : \mathbb {R}^{2} \to \mathbb {R}^{2}$ that preserves distances between points, i.e. $d(f(x), f(y)) = d(x, y)$ for all $x, y$.

Here is a sketch of an argument that any rigid motion of the plane is a composition of translations, rotations, and reflections. Starting with a rigid motion $f$, form a new rigid motion $g$ from $f$ by translating by $-f(0)$: $g = T_{-f(0)} f$. Then $g(0) = 0$. Now use the relation of the dot product with the distance function, $\left\langle x - y, x - y \right\rangle = d(x, y)^2$, to show that $g(0) = 0$ and $d(g(x), g(y)) = d(x, y)$ implies that $\left\langle g(x), g(y) \right\rangle = \left\langle x, y \right\rangle$. Thus $g$ will send unit vectors to unit vectors and orthogonal vectors to orthogonal vectors. In particular, $q_1 = g(e_1)$, $q_2 = g(e_2)$ are orthogonal unit vectors. If $Q$ denotes multiplication by the orthogonal matrix with column vectors $q_1, q_2$, then $Q$ is a rotation or reflection, and $h = Q^{-1} g$ is a rigid motion which preserves $0, e_1, e_2$. Then $h$ can be shown to be the identity by using the relation $v = \left\langle v, e_1 \right\rangle e_1 + \left\langle v_2, e_2 \right\rangle e_2$.

Fill in the details of this argument to show that a rigid motion in the plane is the composition of rotations, reflections, and translations.

The unit disk $D^2 = \left\{ (x, y) : x^2 + y^2 \leq 1\right\}$ is homeomorphic to the square $S = \left\{ (x, y) : \left|x\right| \leq 1, \left|y\right| \leq 1\right\}$ via the following geometrically described map. Each ray from the origin intersects $D^2$ and $S$ in a line segment, and the intersection with $D^2$ is sent linearly to the intersection with $S$: given a point $x$ inside the disk which is not the center, form the circle about the center on which $x$ lies, then form the square which circumscribes this circle, and send $x$ to the point $f(x)$ on the intersection of the perimeter of this square and the ray through $x$; the region between two circles is then sent to the region between the corresponding squares.

Here are the relevant geometrical facts about the basic open sets under this correspondence. The basic open sets inside the circle are given by the regions between two circles, lying between two lines of angles $\theta = \theta_1$, $\theta = \theta_2$, as well as disks about the center. For the inside of the square, the basic open sets are given by regions between two smaller squares, again limited by the same two radial lines, as well as small squares about the center. The map above gives a correspondence between these basic open sets about $x$ and $f(x)$; at the center, a small disk about the center corresponds to a small rectangle about the center.

Use these geometrical facts to verify that this construction gives a homeomorphism. You will need to use the fact that any open set about a point contains one of the basic open sets as described above.

Write down a formula for a homeomorphism between the unit disk $D^2$ and the ellipse $E = \left\{ (x, y) : x^2 + y^2/4 \leq 1\right\}$, and check whether it satisfies $f(tx, ty) = t f(x, y)$, $0 \leq t \leq 1$.

A set $A$ of a metric space is said to be bounded if it is contained in some ball $B(x, r)$. Show that a subset of a metric space which is compact must be bounded.

Show that a finite union of compact sets is compact.

Decide whether or not the following subsets of $\mathbb {R}$ are compact:

A topological space $X$ is called Hausdorff if given $x, y \in X$ there are disjoint open sets $U_x, U_y$ with $x \in U_x$, $y \in U_y$.

Show that in a Hausdorff space, compact sets are closed.

Consider the function $f : [0, 1) \to S^1$ given by $f(x) = (\cos 2\pi x, \sin 2\pi x)$.

Show that $f$ is a bijection that is continuous but is not a homeomorphism.

Analogous to the definition of least upper bound is that of greatest lower bound. Give a definition of greatest lower bound for a set $A \subset \mathbb {R}$ and use the least upper bound property to show that a set with a lower bound must have a greatest lower bound.

Give an example of a closed, bounded subset $A$ of a metric space $X$ that is not compact.

Suppose that $X$ and $Y$ are topological spaces and consider their product $X \times Y = \left\{ (x, y) : x \in X, y \in Y\right\}$. Define a set $W \subset X \times Y$ to be open if given any $(x, y) \in W$, then there are open sets $U$ in $X$ and $V$ in $Y$ so that $(x, y) \in U \times V \subset W$.

Verify that open sets as defined above satisfy the three properties required of a topology.

Proposition 1.5.4: Suppose $X$ is a metric space and $x \in X$. If $\left\{ s_n\right\}$ is a sequence so that for any ball about $x$, the ball contains an infinite number of the $s_n$ (this means that there are an infinite number of $n$ so that $s_n$ is in the ball), then there is a subsequence which converges to $x$.

The proof begins as follows: choose $n_1$ so that $s_{n_1}$ is contained in the ball of radius $1$ about $x$. Since there are an infinite number of the $s_n$ in the ball of radius $\tfrac {1}{2}$ about $x$, we can find $n_2$ so that $n_2 > n_1$ and $s_{n_2} \in B(x, \tfrac {1}{2})$. Inductively, we then use the same idea to choose $n_1 < n_2 < n_3 < \cdots$ so that $s_{n_j} \in B(x, 1/j)$. This will give a convergent subsequence.

Fill in the details in this proof: show that the resulting subsequence $\left\{ s_{n_j}\right\}$ does converge to $x$.

Show that if $\left\{ s_n : n \in \mathbb {N}\right\}$ is finite, then the sequence has a convergent subsequence.

Let $(X, d_X)$, $(Y, d_Y)$ be metric spaces. Then $f : X \to Y$ is said to be uniformly continuous if given $\epsilon > 0$ there exists a $\delta > 0$ such that for $x_1, x_2 \in X$, $d_X(x_1, x_2) < \delta$ implies $d_Y(f(x_1), f(x_2)) < \epsilon$.

Show that $f$ uniformly continuous implies $f$ is continuous, but construct an example to show that the converse does not hold.

Let $f : X \to Y$ be a continuous map of the compact metric space $(X, d_X)$ to the metric space $(Y, d_Y)$. Show that $f$ is uniformly continuous.

Show that a space $X$ is connected iff the only subsets of $X$ which are both open and closed are $\emptyset$ and $X$.

Proposition 1.6.1:

Consider the property: $(*)$ if $x, y \in A \subset \mathbb {R}$, then the interval $[x, y] \subset A$.

Proposition 1.6.3: A set $A \subset \mathbb {R}$ satisfies $(*)$ iff it is an interval, a ray, or $\mathbb {R}$.

It is straightforward to see that an interval, ray, or $\mathbb {R}$ satisfies $(*)$. For the converse, suppose $A$ satisfies $(*)$. The case where $A$ is bounded both from above and below has already been handled: let $a$ be the greatest lower bound and $b$ the least upper bound of $A$. This implies $A \subset [a, b]$. To show that $(a, b) \subset A$: let $c$ be a point in $(a, b)$. Since $a$ is the greatest lower bound, there is an element $e \in A$ with $a \leq e < c$. Similarly, $b$ being the least upper bound implies that there is an element $f \in A$ with $c < f \leq b$. But $(*)$ implies that $[e, f] \subset A$ and so $c \in A$. Hence $(a, b) \subset A$. But $A \subset [a, b]$, so there are four possibilities for $A$: $(a, b)$, $[a, b)$, $(a, b]$, $[a, b]$, each of which is an interval.

Complete the proof of the proposition by considering the other cases: where $A$ is not bounded on one side or the other or both.

Show that $\mathbb {R}$ is connected.

Proposition 1.6.6: Suppose that $A_i$ is a collection of connected subsets of a topological space $X$ so that they all have at least one point $a$ in common. Then the union $A = \cup_i A_i$ is connected.

Use this proposition to deduce that $\mathbb {R}$ is connected from the fact that a closed interval is connected.

A space $X$ is called path connected if, given $x, y \in X$, there is a continuous map $f : [0, 1] \to X$ (called a path in $X$) with $f(0) = x$ and $f(1) = y$. We say that the path connects $x$ to $y$.

Define a relation on $X$ by $x \sim y$ if there is a path connecting $x$ to $y$.

Show that the relation $x \sim y$ as defined above is an equivalence relation.

Show that a union of path connected sets with a point in common is path connected.

The intermediate value theorem states: suppose that $f : X \to \mathbb {R}$ is a continuous function and $X$ is connected. Let $c = f(x_1)$ and $d = f(x_2)$ and suppose that $c < e < d$. Then there is $x_3 \in X$ with $f(x_3) = e$.

Prove the intermediate value theorem using the fact that $f(X)$ is connected and the characterization of connected sets in the line (a subset of $\mathbb {R}$ is connected iff it is an interval, a ray, or $\mathbb {R}$).

Show that the letter T is not homeomorphic to the letter O.

Show that $S^1$ is not homeomorphic to $\mathbb {R}$ by showing $S^1 \setminus \left\{ x\right\}$ is not homeomorphic to $\mathbb {R} \setminus \left\{ y\right\}$.

Show that two disjoint concentric circles in the plane are not homeomorphic to one circle.

Suppose $q : X \to Y$ and $Y$ has the quotient topology with respect to $(X, q)$ (i.e. $U \subset Y$ is open iff $q^{-1}(U) \subset X$ is open). Show $C \subset Y$ is closed iff $q^{-1}(C) \subset X$ is closed.

Consider $X = [0, 1]$ with $Y$ the quotient space formed by identifying $0$ with $1$, and $X' = \mathbb {R}$ with $Y'$ the quotient space formed by identifying $x$ with $x + n$ for all $n \in \mathbb {Z}$. Let $q : X \to Y$, $q' : X' \to Y'$ be the identification maps, and let $i : X \to X'$ be the inclusion. Since $[i(0)] = [0] = [1] = [i(1)]$, $i$ induces a map $\bar{i} : Y \to Y'$ defined by $\bar{i}([x]) = [i(x)]$, which is a continuous bijection onto $Y'$.

Construct an inverse for $\bar{i}$ and show that it is continuous.

Describe basic open sets in the quotient space $[0, 1] \times [0, 1] / (0, t) \sim (1, t), (s, 0) \sim (s, 1)$ about each of the points $[(0, 0)]$, $[(\tfrac {1}{2}, 0)]$, $[(0, \tfrac {1}{2})]$, and $[(\tfrac {1}{2}, \tfrac {1}{2})]$. Describe the inverse image $q^{-1}(U)$ of each of these basic open sets.

Show that the quotient space formed from a square by identifying all of the points in the bottom edge of the square to each other is homeomorphic to a triangle.

Triangle as a quotient space of the square.

Let $A_i \cup_{f_i} B_i = A_i \sqcup B_i / x \sim f_i(x), x \in K_i \subset B_i$ be the quotient space of $A_i \cup B_i$ coming from identifying $x \in K_i \subset B_i$ with $f(x) \in A_i$ via a homeomorphism $f_i : K_i \to f(K_i)$, $i = 1, 2$.

Proposition 1.7.6: Suppose there are homeomorphisms $F_A : A_1 \to A_2$, $F_B : B_1 \to B_2$ with $F_B(K_1) = K_2$ and $F_A f_1 = f_2 F_B$. Then the map $F : A_1 \cup_{f_1} B_1 \to A_2 \cup_{f_2} B_2$ given by $F(x) = F_A(x)$ if $x \in A_1$ and $F(x) = F_B(x)$ if $x \in B_1$ is a homeomorphism.

Prove Proposition 1.7.6.

Consider the space $X$ formed from two copies of $R = D^1 \times D^1$ by identifying $\left\{ -1, 1\right\} \times D^1$ to itself via $d$ with $d(1, y) = (1, -y)$ and $d(-1, y) = (-1, -y)$; that is, $X = R \cup_d R$. Construct a homeomorphism between $X$ and the annulus.

Suppose $X = A \cup_g B$, $Y = A \cup_{g'} B$, where $g, g' : K \subset B \to g(K), g'(K) \subset A$ are homeomorphisms. Suppose $(g')^{-1} g : K \to K = h|_K$, where $h : B \to B$ is a homeomorphism. Show that the identity on $A$ and $h$ on $B$ piece together to give a homeomorphism from $X$ to $Y$.

Show that the Möbius band can also be described as a quotient space $D^1 \times D^1 \cup_f D^1 \times D^1$, with $f(-1, t) = (-1, t)$, $f(1, t) = (1, -t)$.

Identify all points in the lower half of the circle to each other. Show that the resulting quotient space $S^1 / \sim$ is homeomorphic to $S^1$.

Put an equivalence relation on the unit disk by making all points on the boundary circle equivalent to each other. Show that the resulting quotient space $D^2 / \sim$ is homeomorphic to $S^2$.

A collection $\mathcal{B} = \left\{ B_i : i \in I\right\}$ of subsets of $X$ is called a basis and its elements are called basis elements if the following properties are satisfied:

Show that the set of balls $\left\{ B(x, r), x \in X, r > 0\right\}$ is a basis for the topology of a metric space $X$; that is, show that it is a basis and the topology it determines is the metric topology.

Show that any open set in $T_{\mathcal{B}}$ (the topology determined by a basis $\mathcal{B}$) is a union of basis elements.

Suppose the topology for $Y$ is determined by a basis. Show that $f : X \to Y$ is continuous iff, for each basis element $B$ of $Y$, $f^{-1}(B)$ is open.

Show that the open intervals give a basis for the topology of $\mathbb {R}$.

Combine the last two exercises to show that a map to $\mathbb {R}$ is continuous iff, for each open interval $I \subset \mathbb {R}$, we have $f^{-1}(I)$ is open. Formulate and prove an analogous statement for maps to a metric space in terms of balls.

Let $X$ be a metric space with metric $d$ and let $A$ be a subset of $X$. Show that the metric topology on $A$ given by $d$ is the same as the subspace topology.

Suppose $A \subset B \subset X$, and $B$ has the subspace topology. Show that, if $A$ is open in $B$ and $B$ is open in $X$, then $A$ is open in $X$.

If $x, y \in \mathbb {R}^{2}$, let $d(x, y) = \left|x_1 - y_1\right|$. Which of the three properties of a metric does $d$ satisfy?

On $\mathbb {R}^{2} \setminus \left\{ 0\right\}$, write each point in polar coordinates as $(r, \theta )$, where $0 \leq \theta < 2\pi$. Define $d((r_1, \theta_1), (r_2, \theta_2)) = \left|r_1 - r_2\right| + \left|\theta_1 - \theta_2\right|$. Show that this gives a metric on $\mathbb {R}^{2} \setminus \left\{ 0\right\}$ but the topology formed is not the usual topology.

A point $x$ is called a limit point of a set $A$ if every open set $U$ containing $x$ intersects $A \setminus \left\{ x\right\}$ in a nonempty set; that is, $U \cap (A \setminus \left\{ x\right\} ) \neq \emptyset$ for $U$ open, $x \in U$. Denote the limit points of $A$ by $A'$.