Week 4 answers to exercises

Week 4 Exercises — solutions

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Exercise 4.1 (basic test of openness). Suppose that $B$ is a base of a topology on $X,$ and call the subsets of $X$ which are members of $B$ basic open sets.

Let $A$ be a subset of $X .$ Prove that the following are equivalent:
- 1. $A$ is open in $X .$
- 2. $A$ is a union of a collection of basic open sets.
- 3. For each point $x \in A,$ there exists a basic open set $U$ such that $x \in U$ and $U \subseteq A .$

Answer to E4.1. [These exercises without answers]

1. $⟺$ 2. by definition of a base.

Proof that 2. $\Rightarrow$ 3.: assume that $A = ⋃ C$ where $C$ is a collection of basic open sets. Let $x \in A .$ By definition of union of a collection of sets, $x$ is contained in at least one set in the collection $C;$ call this set $U .$ The choice of $U$ ensures that

• $x \in U;$
• $U \in C,$ and $C$ is a collection of basic open sets, so $U$ is a basic open set;
• $U \in C,$ and $⋃ C = A,$ so $U \subseteq A .$

We have proved that 3. holds.

Proof that 3. $\Rightarrow$ 2.: assume that 3. holds. For each point $x \in A,$ 3. allows us to choose a basic open set $U_{x}$ such that $x \in U_{x} \subseteq A .$

We claim that the union of the collection ${U_{x}}_{x \in A}$ of basic open sets is $A .$ Indeed,

• for all $y \in A,$ we have $y \in U_{y}$ by the choice of $U_{y},$ hence $y \in ⋃_{x \in A} U_{x};$ this proves that $A \subseteq ⋃_{x \in A} U_{x};$
• for each $x \in A,$ $U_{x} \subseteq A,$ and so $⋃_{x \in A} U_{x} \subseteq A .$

Thus $⋃_{x \in A} U_{x} = A,$ and so 2. holds.

Exercise 4.2 (the Euclidean topology has a countable base). Consider the Euclidean space $R^{2},$ and let $Q$ be the (countable) collection of all open squares in $R^{2}$ where the coordinates of all four vertices are rational numbers. Prove that $Q$ is a base for the Euclidean topology.

Deduce that the collection of all open sets in the Euclidean space $R^{2}$ has cardinality $ℵ$ (continuum), whereas the collection of all subsets of $R^{2}$ has cardinality $2^{ℵ} .$

Answer to E4.2. [These exercises without answers]

Denote by $Q_{(a, b)}^{s \times s}$ the square of size $s \times s$ whose bottom left corner is the point $(a, b)$ in $R^{2} .$

1. First, we show that every square $Q_{(a, b)}^{s \times s}$ with $a, b, s$ real is a union of some collection ${Q_{(a_{n}, b_{n})}^{s_{n} \times s_{n}}}_{n \geq 1}$ of squares with $a_{n}, b_{n}, s_{n}$ rational.

Indeed, let $(a_{n})$ be a sequence of rational numbers such that $a_{n} \geq a$ and $lim_{n \to \infty} a_{n} = a .$ Also, let $(s_{n})$ be a sequence of rational numbers such that $lim s_{n} = s$ and $a_{n} + s_{n} \leq a + s .$ (It is not difficult to show that such sequences of rational numbers exist.)

Then $Q_{(a_{n}, b_{n})}^{s_{n} \times s_{n}} \subseteq Q_{(a, b)}^{s \times s}$ for all $n,$ and $⋃_{n \geq 1} {Q_{(a_{n}, b_{n})}^{s_{n} \times s_{n}}} = Q_{(a, b)}^{s \times s} .$ See the Figure for illustration.

(a dashed square with nested muticolored dashed squares inside)

2. Now we argue that every set which is open in the Euclidean plane $R^{2}$ is a union of some open squares. Recall that an open square plays the role of $d_{\infty}$ -open ball where the metric $d_{\infty}$ on $R^{2}$ is defined by

$d_{\infty} ((x_{1}, x_{2}), (y_{1}, y_{2})) = max (| x_{1} - y_{1} |, | x_{2} - y_{2} |),$

see the discussion after Proposition 2.3. Specifically, $Q_{(a, b)}^{s \times s}$ is the open ball $B_{r}^{d_{\infty}} ((a + \frac{s}{2}, b + \frac{s}{2}))$ where $r = \frac{s}{2} .$

By definition of metric topology, open balls form a base of topology so it follows that every $d_{\infty}$ -open set in $R^{2}$ is a union of squares, and by 1., a union of rational squares.

It remains to recall that “ $d_{\infty}$ -open” is the same as “Euclidean open”, because the metric $d_{\infty}$ is Lipschitz equivalent to the Euclidean metric $d_{2},$ see Proposition 2.3.

Exercise 4.3 (subbase). Let $(Y, T)$ be a topological space. A subbase of $T$ is a collection $S$ of open sets such that finite intersections of sets from $S$ form a base of $T .$

It is worth noting that, given any set $Y$ (without topology) and any collection $S$ of subsets of $Y,$ we can construct a topology $T_{S}$ on $X$ by using $S$ as a subbase. That is, $T_{S}$ consists of arbitrary unions of finite intersections of members of $S .$ It is not difficult to show that this collection $T_{S}$ is a topology.

Prove that the collection of all open rays in the real line, i.e., sets of the form $(- \infty, a)$ and $(b, + \infty),$ is a subbase of the Euclidean topology.

Answer to E4.3. [These exercises without answers]

Let $S$ be the collection of all open rays in $R .$ By taking intersections of just two sets from $S,$ we can generate all open bounded intervals in $R :$

$(b, a) = (- \infty, a) \cap (b, + \infty) .$

Since the open intervals $(b, a),$ where $a, b \in R,$ form a base of the Euclidean topology on $R,$ the topology $T_{S}$ generated by the subbase $S$ contains the Euclidean topology.

On the other hand, every set in $S$ is open in the Euclidean topology, hence so are unions of finite intersections of sets from $S .$ Therefore, the topology $T_{S}$ is contained in the Euclidean topology.

We conclude that $T_{S}$ is equal to the Euclidean topology, as claimed.

Exercise 4.4 (subbasic test of continuity). Let $X,$ $Y$ be topological spaces, $f : X \to Y$ be a function, and $S$ be a subbase of topology on $Y .$ Prove that the following are equivalent:
- 1. $f$ is continuous.
- 2. The preimage of every subbasic set in $Y$ is open in $X$ (meaning: $\forall V \in S,$ $f^{- 1} (V)$ is open in $X .$ )

Answer to E4.4. [These exercises without answers]

1. $\Rightarrow$ 2.: by definition of subbase, $S$ is a subcollection of the topology on $Y,$ i.e., every subbasic set in is open in $Y .$ By definition of “continuous”, the preimage of an open set is open, and so the preimages of subbasic sets must be open in $X,$ proving 2.

2. $\Rightarrow$ 1.: a base $B$ of topology on $Y$ consists of sets of the form $V_{1} \cap \dots \cap V_{n},$ where $n \geq 0$ and $V_{1}, \dots, V_{n} \in S .$ The preimage of intersection is the intersection of preimages, so we have

$f^{- 1} (V_{1} \cap \dots \cap V_{n}) = f^{- 1} (V_{1}) \cap \dots \cap f^{- 1} (V_{n}),$

and, since $f^{- 1} (V_{i})$ is open in $X$ by 2., and a finite intersection of open sets is open, we conclude that $f^{- 1} (V)$ is open in $X$ for all $V \in B .$

Finally, every open set in $Y$ is a union of sets from $B,$ and the preimage of a union is the union of preimages. We conclude that $f^{- 1} ($ open set in $Y)$ is open in $X,$ hence, by definition of “continuous”, $f$ is continuous, proving 1.

Exercise 4.5. (a) Let $X$ be a topological space and let $f : X \to R$ be a function. Prove: $f$ is continuous iff for all $a, b \in R,$ the sets $X_{f < a} = {x \in X : f (x) < a}$ and $X_{f > b} = {x \in X : f (x) > b}$ are open in $X .$

(b) Let $X$ be a topological space and let $f, g : X \to R$ be continuous functions. Prove that the function $f + g : X \to R$ is continuous. Hint: use (a).

Answer to E4.5. [These exercises without answers]

(a) Note that $X_{f < a} = f^{- 1} ((- \infty, a))$ and $X_{f > b} = f^{- 1} ((b, + \infty)) .$ The sets $(- \infty, a)$ and $(b, + \infty)$ form a subbase of the Euclidean topology on $R$ (see an earlier exercise). Hence by the subbasic test of continuity (see the previous exercise), $f$ is continuous iff all the sets $X_{f < a}$ and $X_{f > b}$ are open.

(b) We need to prove that the sets $X_{f + g < a},$ $X_{f + g > b}$ are open for all $a, b \in R .$ Note that

$f (x) + g (x) < a ⟺ \exists t \in R : f (x) < t, g (x) < a - t .$

Indeed, $\Leftarrow$ is obvious, and to see $\Rightarrow,$ take $t$ to be any real number in the interval $(f (x), a - g (x)) .$ The above rewrites in terms of sets as

$X_{f + g < a} = ⋃_{t \in R} X_{f < t} \cap X_{g < a - t} .$

Since $f, g$ are continuous, by (a) the sets $X_{f < t}$ and $X_{g < a - t}$ are open in $X;$ the intersection of two open sets is open, and the union of any collection of open sets is open, which shows that $X_{f + g < a}$ is open.

It is shown in the same way that $X_{f + g > b}$ is an open subset of $X,$ for all $b \in R .$ We now use (a) again to conclude that $f + g$ is a continuous function.

Version 2024/11/26. These exercises in PDF To other course material