Abstract Nonsense

Crushing one theorem at a time

Just For Fun (Rudin’s Topology Section)

Just for fun I’ve decided to do the entire section on topology in Rudin’s Principles of Mathematical Analysis. The problems themselves aren’t particularly difficult, but some are interesting and it will give me a chance to do some refreshing. Also, a friend of mine is starting Rudin and it will be nice to have done them in case he has any questions. I will probably prove more general results from topology if I feel like it. So…


Problem: Prove that the empty set is a subset of every set.

Proof: Suppose that we are working in some universal set U and there existed some E\subseteq U such that \varnothing\not\subseteq E. Then, by definition there exists some x\in \varnothing such that x\notin E. But, this is clearly preposterous since the statement x\in \varnothing is never true. \blacksquare


Problem: A complex number z is said to be algebraic if there are integers a_1,\cdots,a_n, not all zero, such that


Let \mathbb{A} denote the set of all algebraic numbers. Prove that \text{card }\mathbb{A}=\aleph_0.

Proof: Let \Omega_n(a_1,\cdots,a_n)=\left\{z\in\mathbb{C}:a_1+\cdots+a_nz^n=0\right\}. By the fundamental theorem of algebra we have that \text{card }\Omega_n(a_1,\cdots,a_n)\leqslant n. So, clearly \displaystyle \bigcup_{(a_1,\cdots,a_n)\in\mathbb{Z}^n}\Omega_n(a_1,\cdots,a_n) being the countable union of countable sets is countable. Thus, we claim that

\displaystyle \mathbb{A}\subseteq\bigcup_{n\in\mathbb{N}}\bigcup_{(a_1,\cdots,a_n)\in\mathbb{Z}^n}\Omega_n(a_1,\cdots,a_n)

from where it will follows that \mathbb{A} being a subset of a countable union of countable sets is countable. So, let z\in\mathbb{A} then there exists a_1,\cdots,a_m\in\mathbb{Z}^m such that a_1+\cdots+a_mz^m=0. Thus, clearly z\in\Omega_m(a_1,\cdots,a_m)\subseteq from where the conclusion follows. \blacksquare


Problem: Prove that there exists real numbers which aren’t algebraic.

Proof: Clearly since \mathbb{A} is countable so is \mathbb{R}\cap\mathbb{A}=\mathbb{R}_{\mathbb{A}}. And thus, if \mathbb{R}-\mathbb{R}_{\mathbb{A}} were countable then so would \mathbb{R}_{\mathbb{A}}\cup\left(\mathbb{R}-\mathbb{R}_{\mathbb{A}}\right)=\mathbb{R} which is of course absurd. It follows that there are uncountably many non-algebraic (transcendental) numbers. \blacksquare


Problem: Is the set of all irrational numbers countable?

Proof: Of course not. For using a similar argument to the above, if \mathbb{I}=\mathbb{R}-\mathbb{Q} were countable then  \mathbb{Q}\cup\mathbb{I}=\mathbb{R} would be as well. \blacksquare


Problem: Construct a bounded set of real numbers with exactly three limit points.

Proof: Why not do it for an arbitrary natural number. Let \displaystyle E_0=\{0\}\cup\left\{\frac{1}{n}:n\in\mathbb{N}\right\}. Evidently E_0 has precisely one limit point. Now, let E_m=\left\{e+m:e\in E_0\right\} then it is equally clear E_m has exactly one limit point (namely m). So, the set K_{m}=E_0\cup\cdots\cup E_{m-1} has precisely m limit points as one can easily check. \blacksquare


Problem: Let D(E) be the set of all limit points of a set E\subseteq\mathcal{M} (where \mathcal{M} is a metric space). Prove that a) D(E) is closed , b) D(E)=D\left(\overline{E}\right), c) Do E and D(E) always have the same limit points?


a) This is true in any T_1 topological space. But, first we need a quick lemma

Lemma: Let X be a T_1 space, and U\subseteq X. Then, if x\in D(U) then for very neighborhood V of x we have that V\cap U is infinite.

Proof: Suppose not, and there existed a neighborhood V of x such that V\cap U=\{u_1,\cdots,u_m\}. Since X is T_1 for each u_k there exists some neighborhood N_k of x such that u_k\notin N_k. Thus, V\cap N_1\cap\cdots\cap N_m is a neighborhood of x which does not intersect U, which of course contradicts that x\in D(U). \blacksquare

So, using this let X be a T_1 space, then for any E\subseteq X it is true that D(E) is closed. To see this let x\in D(D(E)) and let N be any neighborhood of it. By assumption there exists some y\in N\cap D(E) and thus since N is a neighborhood of y also there exists infinitely many points of E in N. In particular there exists some e\in N\cap E such that e\ne x. Since N was arbitrary it follows that x\in D(E). The conclusion follows.

b) Let x\in D(E) then every neighborhood N contains a point of E different from itself, and thus a point of \overline{E} different from itself. Conversely, let x\in D(\overline{E})  and let N be any neighborhood of it. We must have that there is some y\in N\cap \overline{E}. Now, if y\in E we’re done, and if not by the previous lemma since y is a limit point of E it must be that N contains infinitely many points of E anyways, and thus a point of E different from x. The conclusion follows.

c) Of course not. \left\{0\right\}\cup\left\{\frac{1}{n}:n\in\mathbb{N}\right\}\subseteq\mathbb{R} with the usual topology has exactly one limit point, namely 0. But, a singleton in Euclidean space does not have an limit points.


Problem Let \mathcal{M} be a metric space and E_1,\cdots,\subseteq\mathcal{M}. a) Show that \displaystyle \overline{\bigcup_{m=1}^{n}E_m}=\bigcup_{m=1}^{n}\overline{E_m}.  b) show that example that \displaystyle \overline{\bigcup_{m=1}^{\infty}E_m}\supseteq\bigcup_{m=1}^{\infty}\overline{E_m} and the inclusion can be strict.


a) This is true in any topological space X. Suppose x\notin\overline{E_1\cup\cdots\cup E_n} then there exists a neighborhood N of x such that N\cap\left(E_1\cup\cdots\cup E_n\right)=(N\cap E_1)\cup\cdots\cup (N\cap E_n)=\varnothing and thus N\cap E_k=\varnothing,\text{ }k=1,\cdots,n and thus x\notin \overline{E_k},\text{ }k=1,\cdots,n and so x\notin\overline{E_1}\cup\cdots\cup\overline{E_n}. Conversely, let x\notin\overline{E_1}\cup\cdots\cup\overline{E_n}. Then, since x\notin \overline{E_k},\text{ }k=1,\cdots,n there exists some neighborhood U_k of it such that U_k\cap E_k=\varnothing. Thus, clearly U_1\cap\cdots\cap U_n is a neighborhood of x which does not intersect E_1\cup\cdots\cup E_n and thus x\notin\overline{E_1\cup\cdots\cup E_n}. The conclusion follows.

b) Let \displaystyle x\notin\overline{\bigcup_{m=1}^{\infty}E_m} then there exists a neighborhood N of x such that \displaystyle N\cap\bigcup_{m=1}^{\infty}E_m=\bigcup_{m=1}^{\infty}\left(N\cap E_m\right)=\varnothing and thus N\cap E_m=\varnothing,\text{ }m\in\mathbb{N}. It follows that x\notin \overline{E_m},\text{ }m\in\mathbb{N} and thus \displaystyle x\notin\bigcup_{m=1}^{\infty}\overline{E_m}. But, as was noted the inclusion can be strict. Give \mathbb{R} the usual topology. Then, for each q\in\mathbb{Q} we have the singleton \{q\} is closed and thus \overline{\{q\}}=\{q\}. So,

\displaystyle \bigcup_{q\in\mathbb{Q}}\overline{\{q\}}=\bigcup_{q\in\mathbb{Q}}\{q\}=\mathbb{Q}\subsetneq\overline{\bigcup_{q\in\mathbb{Q}}\{q\}}=\overline{\mathbb{Q}}=\mathbb{R}


Problem: a) Is every point of every open set E\subseteq\mathbb{R}^2 a limit point of E? Answer the same for closed sets in \mathbb{R}?


a) Of course, I mean this is true in any topological space. If E is open then each point of E is an interior point and thus trivially a limit point.

b) No, take a singleton.


Problem: Let E^{\circ} denote the set of all interior points for E in a metric space \mathcal{M}. Prove that:

a) E^{\circ} is always open

b) E^{\circ} is open if and only if E^{\circ}=E.

c) If G\subseteq E and G is open then G\subseteq E^{\circ}.

d) \mathcal{M}-E^{\circ}=\overline{\mathcal{M}-E}

e) Do E and \overline{E} always have the same interiors?

f) Do E and E always have the same interiors?


a) Let x\in E^{\circ} then there exists a neighborhood N of x such that N\subseteq E, but for each y\in N we also have that N is a neighborhood of it which is contained in E. Thus, N\subseteq E^{\circ}.

b) If E=E^{\circ} the above proves that E is open. Conversely, since a set in a metric space is open if and only if each point is an interior point the converse is trivial.

c) Let x\in G then G is automatically a neighborhood of x which is contained in E so that x\in E^{\circ}.

d) Let x\in\mathcal{M}-E^{\circ} then for every neighborhood N of x we have that N\not\subseteq E and so N\cap \mathcal{M}-E\ne\varnothing and thus x\in\overline{\mathcal{M}-E}. Conversely, let x\in\overline{\mathcal{M}-E} then for every neighborhood N of x we have that N\cap\mathcal{M}-E\ne\varnothing and thus N\not\subseteq E and so x\notin E^{\circ} or x\in\mathcal{M}-E^{\circ}.

e) Of course not. It is trivial to check that if one gives \mathbb{R} the usual topology then \mathbb{Q}^{\circ}=\varnothing (since it’s complement is dense in \mathbb{R}) and \overline{\mathbb{Q}}=\mathbb{R}.

f) The above works as a counterexample to this as well.


Problem: Let \mathcal{M} be an infinite set and define d:\mathcal{M}\times\mathcal{M}\to\mathbb{R} by

d(x,y)=\begin{cases} 0 \quad\text{if}\quad x=y \\ 1\quad\text{if}\quad x\ne y \end{cases}

a) Prove that this is a metric. b) Which subsets of the resulting metric space are open? c) Which are closed? d) Which are compact?


a) Clearly d(x,y)\geqslant 0 and d(x,y)=0\Leftrightarrow x=y. Also, since equality and inequality are transitive symmetry immediately follows. The triangle inequality is just a set of three cases. If x=y=z then d(x,y)=0\leqslant d(x,z)+d(y,z)=0+0. If x=z,y\ne z then d(x,y)=1\leqslant d(x,z)+d(y,z)=1+0=1. If x\ne z,y\ne z then d(x,y)\leqslant 1\leqslant d(x,z)+d(y,z)=1+1.

b) For every x\in\mathcal{M} we have that \{x\} is open since B_{\frac{1}{2}}(x)=\{x\}. Thus, all subsets of \mathcal{M} being the union of singletons are open.

c) From b) it readily follows that every subset of \mathcal{M} is also open.

d) E\subseteq\mathcal{M} will be compact if and only if E is finite. To see this suppose that E is infinite, then \left\{\{e\}\right\}_{e\in E} is evidently an open cover of it which admits no finite subcover. The converse is trivial.


Problem: Let the following be maps from \mathbb{R}^2 to \mathbb{R}, and determine if they determine a metric. a) d(x,y)=(x-y)^2, b) d(x,y)=\sqrt{|x-y|}, c) d(x,y)=|x^2-y^2|, d) d(x,y)=|x-2y|, e) \displaystyle d(x,y)=\frac{|x-y|}{1+|x-y|}


a) This is not a metric. (4-1)^2=9>5=4+1=(4-2)^2+(1-2)^2

b) This is a metric as one (I’m to lazy to tex it) can check.

c) Nope, d(1,-1)=0.

d) Negatory, d(1,\tfrac{1}{2})=0.

e) Yes. In fact, (once again I omit the proof) if d is a metric so is \displaystyle \frac{d}{1+d}


Problem: Let K\subseteq\mathbb{R} with the usual topology be defined by K=\{0\}\cup\left\{\frac{1}{n}:n\in\mathbb{N}\right\}. Prove directly (i.e. no Heine-Borel) that K is compact.

Proof: This is true in any topological space X. But, for the sake of convenience let us do it merely in a Hausdorff space X. So, let \{x_n\}_{n\in\mathbb{N}} is convergent sequence in X with x_n\to x. Then, C=\{x\}\cup\left\{x_n:n\in\mathbb{N}\right\} is a compact subspace of X.

Proof: Let \Omega be an open cover for C. Choose U\in\Omega such that x\in U. Then, since x_n\to x all but finitely may values of \{x_n:n\in\mathbb{N}\} lie in U. So, for each of the other finitely many points of C-U choose any element of \Omega which contains them. Clearly this will be a finite subcover. \blacksquare


May 13, 2010 - Posted by | Analysis, Fun Problems, Topology, Uncategorized | , , , ,


  1. Could you please make email contact with me? I’d like to pay you to tutor me via video chat. As you can tell from my email address, I am either a student at UA or a spambot disguised as a student! Thanks,

    Comment by C W | October 7, 2010 | Reply

    • I’m sorry friend, that doesn’t sound like a very good idea. I am way too busy this term.

      Comment by drexel28 | October 7, 2010 | Reply

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