Topology and Functional Analysis Questions and Answers

Answer: B

The statement that is not always true is Union of any family of closed sets is closed.

Only finite unions of closed sets are guaranteed closed. In \(\mathbb{R}\), the closed sets \([1/n, 1]\) for \(n = 1, 2, \ldots\) have union \((0, 1]\), which is not closed.

The other three always hold. A finite union of compact sets is compact, since a finite collection of finite subcovers is still finite. Connected sets sharing a common point have a connected union, because any separation would have to split one of the sets. A union of dense subsets contains a dense subset, and any superset of a dense set is dense.

Answer: C

The space is normal and connected, so the answer is R and S.

The only open set containing \(c\) is \(X\) itself. So \(a\) and \(c\) cannot be separated by disjoint open sets, and the space is not Hausdorff (P fails).

The closed sets are \(X, \{b, c\}, \{a, c\}, \{c\}\) and \(\phi\). The point \(a\) and the closed set \(\{b, c\}\) cannot be separated, because any open set containing \(\{b, c\}\) is \(X\); so the space is not regular (Q fails).

Every non-empty closed set contains \(c\), so no two non-empty closed sets are disjoint and normality holds vacuously (R). In any split of \(X\) into two disjoint non-empty open sets, one of them would have to contain \(c\) and so be all of \(X\); hence the space is connected (S).

Answer: C

The set is closed in both \((\mathbb{R}, \tau_{1})\) and \((\mathbb{R}, \tau_{2})\).

\(A = \{x : 4\sin^{2} x \le 1\}\) is the preimage of the closed set \((-\infty, 1]\) under the continuous function \(4\sin^{2} x\), so it is closed in the usual topology. Adding the single point \(\frac{\pi}{2}\) (where \(4\sin^{2} x = 4\), so it was not in \(A\)) keeps the set closed, since a finite union of closed sets is closed.

\(\tau_{2}\) is the lower-limit (Sorgenfrey) topology, which is finer than \(\tau_{1}\): every usual open set is a union of intervals \([a, b)\). A set closed in \(\tau_{1}\) has a \(\tau_{1}\)-open complement, which is also \(\tau_{2}\)-open, so the set is closed in \(\tau_{2}\) too.

Answer: C

The topology of \(Z\) is strictly coarser (weaker) than that of \(X\).

A basic open set of \(Z\) is \(U \times V\) with \(U\) and \(V\) co-finite in \(\mathbb{R}\). The complement of a finite set is open in the usual topology, so \(U \times V\) is open in \(\mathbb{R}^{2}\) with the usual topology. Hence every open set of \(Z\) is open in \(X\).

The inclusion is strict. Any two non-empty open sets of \(Z\) intersect, because co-finite sets in \(\mathbb{R}\) always meet, so \(Z\) is not Hausdorff, while \(X\) is. For instance the open unit disc is open in \(X\) but contains no basic set \(U \times V\), each of which is unbounded.

So the two topologies are comparable and not equal, and it is \(Z\), not \(X\), that is coarser.

Answer: B

The true statement is \(\Delta\) is locally compact.

The map x ↦ (x, x, x, ...) from \(\mathbb{R}\) onto \(\Delta\) is continuous, because each coordinate is continuous, and its inverse is the first projection, also continuous. So \(\Delta\) is homeomorphic to \(\mathbb{R}\), which is locally compact (and connected).

\(\Delta\) is not open in Y: every basic open set of the product topology restricts only finitely many coordinates, so it contains points whose coordinates are not all equal. It is not dense: \(\Delta\) is closed, being the set where every coordinate equals the first, and it is not all of Y. It is not disconnected, since \(\mathbb{R}\) is connected.

Answer: D

The correct statement is X is NOT homeomorphic to W.

W, the union of the three coordinate planes, is connected. X is the origin together with countably many octahedral surfaces \(|x|+|y|+|z| = q\), one for each positive rational q. It is not connected: for any irrational r, the sets where \(|x|+|y|+|z| \lt r\) and \(\gt r\) split X into two disjoint non-empty open pieces. Connectedness is preserved by homeomorphisms, so X and W are not homeomorphic.

The other statements fail for similar reasons: Z, the unit sphere, is compact while W is not; Y (xyz = 1) has four connected components while W is connected; and X has infinitely many components (indeed it is not locally connected near the origin) while Y has four.

Answer: C

The statement that need not hold is \(\partial A \subseteq \partial(\text{int}(A))\).

In \(\mathbb{R}\), take \(A = \{0\}\). Its interior is empty, so \(\partial(\text{int}(A)) = \emptyset\), but \(\partial A = \{0\}\). So the inclusion fails.

The others always hold. \(\text{int}(X \setminus A)\) is an open set missing \(A\), so it misses \(\overline{A}\) too. \(A \subseteq \overline{A}\) by definition. And since \(\text{int}(A) \subseteq \text{int}(\overline{A})\),

\[ \begin{aligned} \partial\overline{A} &= \overline{A} \setminus \text{int}(\overline{A}) \\ &\subseteq \overline{A} \setminus \text{int}(A) = \partial A. \end{aligned} \]

Answer: A

Both I and II are TRUE.

This is Furstenberg's topology on \(\mathbb{Z}\). I: \(S(a, b)\) is open as a basis set, and its complement is the union of the other residue classes \(S(a, b + 1), \ldots, S(a, b + |a| - 1)\), which are open; so \(S(a, b)\) is also closed.

II: take any set \(C\) containing \(x\) and another point \(y \ne x\). Choose \(a\) with \(a \nmid (y - x)\); then \(S(a, x)\) is clopen, contains \(x\) and not \(y\), and splits \(C\) into two disjoint relatively open pieces. So no connected set has more than one point. Hence options with I or II false are wrong.

Answer: A

Both I and II are TRUE.

\(\mathbb{R}/\mathbb{Z}\) is homeomorphic to the circle \(S^{1}\) via \(t \mapsto e^{2\pi it}\). I: it is the continuous image of the compact interval [0, 1], so it is compact.

II: removing one point from a circle leaves a space homeomorphic to an open interval, which is connected. So both statements hold, and options declaring either false are wrong.

Answer: A

X is homeomorphic to the interval [-1, 1].

Two points of the sphere are identified exactly when they have the same height \(x_{3}\), so each equivalence class is a horizontal circle of latitude (a single point at each pole). The height map \(h(x_{1}, x_{2}, x_{3}) = x_{3}\) is continuous, onto [-1, 1], and constant precisely on the classes, so it induces a continuous bijection \(X \to [-1, 1]\).

A continuous bijection from a compact space (X is the image of the compact sphere) onto a Hausdorff space is a homeomorphism. The circle, disc and cylinder in the other options would keep more of the sphere's structure than the relation leaves, which collapses each latitude circle to a point.