Week 4 Exercises — solutions

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

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

1.\(\Longleftrightarrow \)2. by definition of a base.

Proof that 2.\(\Rightarrow \)3.: assume that \(A=\bigcup \mathscr C\) where \(\mathscr C\) 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 \(\mathscr C;\) call this set \(U.\) The choice of \(U\) ensures that

• • \(x\in U;\)

• • \(U\in \mathscr C,\) and \(\mathscr C\) is a collection of basic open sets, so \(U\) is a basic open set;

• • \(U\in \mathscr C,\) and \(\bigcup \mathscr C = A,\) 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 \(y\in \bigcup \limits _{x\in A} U_x;\) this proves that \(A\subseteq \bigcup \limits _{x\in A} U_x;\)

• • for each \(x\in A,\) \(U_x\subseteq A,\) and so \(\bigcup \limits _{x\in A} U_x\subseteq A.\)

Thus \(\bigcup \limits _{x\in A} U_x = A,\) and so 2. holds.

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 \(\RR ^2.\)

1. First, we show that every square \(Q_{(a,b)}^{s\times s}\) with \(a,b,s\) real is a union of some collection \(\{Q_{(a_n,b_n)}^{s_n\times s_n}\}_{n\ge 1}\) of squares with \(a_n,b_n,s_n\) rational.

Indeed, let \((a_n)\) be a sequence of rational numbers such that \(a_n\ge a\) 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 \(a_n+s_n\le a+s.\) (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 \(\bigcup _{n\ge 1}\{Q_{(a_n,b_n)}^{s_n\times s_n}\} = Q_{(a,b)}^{s\times s}.\) See the Figure for illustration.

2. Now we argue that every set which is open in the Euclidean plane \(\RR ^2\) 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 \(\RR ^2\) 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 \(B^{d_\infty }_r((a+\frac s2,b+\frac s2))\) where \(r=\frac s2.\)

By definition of metric topology, open balls form a base of topology so it follows that every \(d_\infty \)-open set in \(\RR ^2\) 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.

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

Let \(\mathscr S\) be the collection of all open rays in \(\RR .\) By taking intersections of just two sets from \(\mathscr S,\) we can generate all open bounded intervals in \(\RR :\)

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

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

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

We conclude that \(\mathscr T_{\mathscr S}\) is equal to the Euclidean topology, as claimed.

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

1. \(\Rightarrow \) 2.: by definition of subbase, \(\mathscr S\) 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 \(\mathscr B\) of topology on \(Y\) consists of sets of the form \(V_1\cap \dots \cap V_n,\) where \(n\ge 0\) and \(V_1,\dots ,V_n\in \mathscr S.\) 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 \(V\in \mathscr B.\)

Finally, every open set in \(Y\) is a union of sets from \(\mathscr B,\) 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.

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 \(\RR \) (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 \(a,b\in \RR .\) Note that

\[ f(x)+g(x)<a \quad \iff \quad \exists t\in \RR : \ f(x)<t, \ g(x)<a-t. \]

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

\[ X_{f+g<a} = \bigcup _{t\in \RR } X_{f<t} \cap X_{g<a-t}. \]

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 \(b\in \RR .\) 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