Your Answer : (Not Answered)

Correct Answer : C

The set must be C = R.

Meeting every open interval means C is dense in R: every real number is a limit of points of C. A closed set contains all its limit points, so \(C = \overline{C} = \mathbf{R}\).

The other options all fail for C = R: it is not compact, not bounded, and not countable. (Density alone would allow a countable set such as Q, but Q is not closed.)

Your Answer : (Not Answered)

Correct Answer : B

The false statement is that every closed set is compact in X.

In the discrete topology every subset is open, and hence every subset is also closed. So 'every open set is closed', 'every closed set is open' and 'every singleton is open' all hold.

Compactness fails as soon as X is infinite: X itself is closed, but the cover of X by its singletons is an open cover with no finite subcover. A subset of a discrete space is compact only when it is finite, so the statement is false for any infinite discrete space.

Your Answer : (Not Answered)

Correct Answer : D

X is regular if each point and each closed set not containing it have disjoint open neighbourhoods.

That is the definition: for every \(x \in X\) and closed B with \(x \notin B\), there are open sets U and V with \(x \in U\), \(B \subset V\) and \(U \cap V = \phi\).

Closed singletons describe the T₁ axiom, a different property. The two options asking only for an open U with \(x \in U\) disjoint from B are too weak: when B is closed, the complement of B is such a U in every space, so the condition holds trivially and separates nothing; when B is open, the condition is not regularity at all.

Your Answer : (Not Answered)

Correct Answer : A

(−1, 1) is homeomorphic to the parabola \(y = x^{2}\).

The parabola is the graph of a continuous function on R, so \(x \mapsto (x, x^{2})\) is a homeomorphism from R onto it. And (−1, 1) is homeomorphic to R, for example by \(x \mapsto \tan\frac{\pi x}{2}\).

The others are distinguished by topological invariants. The ellipse and the circle are compact, while (−1, 1) is not. [−1, 1] is compact too. Removing a point from the circle or ellipse leaves it connected, but removing a point from an open interval disconnects it.

Your Answer : (Not Answered)

Correct Answer : C

\(f([0, 1] \cap Q)\) is dense in S¹.

The map \(f(x) = e^{2\pi i x}\) is continuous from [0, 1] onto the circle. The rationals in [0, 1] are dense in [0, 1], and a continuous surjection carries a dense set onto a dense subset of the image, so the points \(e^{2\pi i q}\), the roots of unity, are dense in \(S^{1}\).

The image is countable, so it cannot be all of \(S^{1}\) or any closed set containing an arc; being dense but not everything, it is not closed. It is not open, because every arc of the circle contains points \(e^{2\pi i x}\) with \(x\) irrational, which are not in the image. And a countable subset of the circle with more than one point is not connected: removing two points with irrational angles, one on each arc between two of its points, separates it into two relatively open pieces.

Your Answer : (Not Answered)

Correct Answer : D

The conclusion is \(|X| \gt 0\): the intersection is non-empty.

Write \(I_{n} = [a_{n}, b_{n}]\). Nesting makes \(a_{n}\) increasing and \(b_{n}\) decreasing, with \(a_{n} \le b_{m}\) for all n, m. So \(a = \sup a_{n}\) exists and satisfies \(a_{n} \le a \le b_{n}\) for every n, which puts a in every \(I_{n}\). This is the nested interval property (Cantor's intersection theorem).

The other options are not forced. X is never empty, so option A fails. X can be a single point (\(I_{n} = [-\frac{1}{n}, \frac{1}{n}]\)) or a whole interval (\(I_{n} = [-1 - \frac{1}{n}, 1 + \frac{1}{n}]\), with intersection [−1, 1]), so neither |X| = 1 nor \(|X| = \infty\) holds in general.

Your Answer : (Not Answered)

Correct Answer : B

The compact set is \(B \setminus A\).

A is the strip \(-1 \le x \lt 1\), \(|y| \le 1\) (the condition \(-1 \le |y|\) is automatic), which is not closed because its right edge \(x = 1\) is missing. B is the closed disk of radius \(1/\sqrt{2} \approx 0.707\) centred at (1, 0).

Every point of B has \(|y| \le 0.707\) and \(x \ge 0.29 \gt -1\), so a point of B fails to lie in A only when \(x \ge 1\). Hence \(B \setminus A = B \cap \{x \ge 1\}\), a closed half-disk: closed and bounded, so compact.

The others are not closed. The points \((1 - 1/n, 0)\) lie in \(A \cap B\) but their limit (1, 0) is not in A, so \(A \cap B\) is not closed. The points \((1 - 1/n, 0.9)\) lie in A but outside B (distance about 0.9 from the centre), and their limit (1, 0.9) lies in neither set, so \(A \setminus B\) and \(A \cup B\) are not closed either.

Your Answer : (Not Answered)

Correct Answer : B

The closure of {0} is \((-\infty, 0]\).

The open sets of this topology are \(\phi\), \(\mathbf{R}\) and the rays \((a, \infty)\), so the closed sets are \(\mathbf{R}\), \(\phi\) and the rays \((-\infty, a]\). The closure of {0} is the smallest closed set containing 0, which is \((-\infty, 0]\).

Check with neighbourhoods: a point \(x \le 0\) is in the closure because every open set containing it, \((a, \infty)\) with \(a \lt x\), contains 0. A point \(x \gt 0\) has the neighbourhood \((0, \infty)\) missing 0. So {0} itself is not closed, \([0, \infty)\) points the wrong way, and \(\mathbf{R}\) is too big.

Your Answer : (Not Answered)

Correct Answer : A

The key's answer is that \(\|\cdot\|_{3}\) and \(\|\cdot\|_{4}\) are norms and complete.

\(\|x\|_{3} = \|Tx\|_{2}\) is a norm on X because T is linear and injective, and \(\|y\|_{4} = \|T^{-1}y\|_{1}\) is a norm on Y because \(T^{-1}\) is linear and injective.

For completeness, T is a bounded bijection between Banach spaces, so by the open mapping (bounded inverse) theorem \(T^{-1}\) is bounded too. Then \(\|x\|_{3} \le \|T\|\,\|x\|_{1}\) and \(\|x\|_{1} \le \|T^{-1}\|\,\|x\|_{3}\), so \(\|\cdot\|_{3}\) is equivalent to the complete norm \(\|\cdot\|_{1}\); likewise \(\|\cdot\|_{4}\) is equivalent to \(\|\cdot\|_{2}\). Equivalent norms share Cauchy sequences, so both are complete, which rules out B, C and D.

Your Answer : (Not Answered)

Correct Answer : A

Y is second countable.

Let \(\{B_{n}\}\) be a countable base for X. Since \(f\) is open, each \(f(B_{n})\) is open in Y. Given an open set \(U \subseteq Y\) and \(y \in U\), pick \(x\) with \(f(x) = y\) (surjectivity); \(f^{-1}(U)\) is open (continuity) and contains \(x\), so some \(B_{n}\) has \(x \in B_{n} \subseteq f^{-1}(U)\). Then \(y \in f(B_{n}) \subseteq U\). So \(\{f(B_{n})\}\) is a countable base for Y.

A second countable space is also first countable and separable, so the options denying either property are wrong.

Your Answer : (Not Answered)

Correct Answer : C

The space that need not be Hausdorff is an indiscrete topological space.

In the indiscrete topology the only open sets are the empty set and X. If X has two points, the only open set containing either of them is X itself, so they cannot be separated by disjoint open sets.

The others are always Hausdorff. A space is Hausdorff exactly when its diagonal \(\Delta\) is closed in \(X \times X\) (A). In a metric space two points at distance \(d\) lie in disjoint balls of radius \(d/2\) (B). And a product of Hausdorff spaces is Hausdorff, since points differing in one coordinate are separated by preimages of disjoint open sets (D).

Your Answer : (Not Answered)

Correct Answer : B

The statement that need not be true is [S] is a closed subspace of X.

The span [S], all finite linear combinations of elements of S, is always a subspace (A). Its closure is again a subspace, because addition and scalar multiplication are continuous (C), and a closure is always a closed set (D).

The span itself need not be closed. In \(l_{2}\), let S be the standard unit vectors \(e_{n}\). Their span is the set of sequences with finitely many non-zero terms. The vector \((1, 1/2, 1/3, \ldots)\) is a limit of such sequences but has infinitely many non-zero terms, so the span is a subspace that is not closed.

Your Answer : (Not Answered)

Correct Answer : C

The true statement is every component of an open set of X is open in X.

This is a standard characterisation: X is locally connected if and only if the components of every open subset are open. If U is open and C is a component of U, each point of C has a connected open neighbourhood inside U, which must lie in C, so C is open.

Local connectedness does not imply connectedness (two disjoint intervals are locally connected but not connected). Nor does it give local path connectedness or make components path components: Kirch's topology on the positive integers is a countable, connected, locally connected Hausdorff space, and a countable Hausdorff space has no non-constant paths, so its single component is not path connected.

Your Answer : (Not Answered)

Correct Answer : D

The map p is neither an open map nor a closed map.

p sends [0, 1/2) to 0 and [1/2, 1] to 1. In the quotient topology a set is open when its preimage is open in [0, 1]. The preimage of {0} is [0, 1/2), open in [0, 1]; the preimage of {1} is [1/2, 1], not open. So \(T_{2}\) = {∅, {0}, {0, 1}}, whose closed sets are {0, 1}, {1} and ∅.

Not open: (1/2, 1] is open in [0, 1] but its image {1} is not open. Not closed: [0, 1/4] is closed in [0, 1] but its image {0} is not closed.

So the options claiming p is open, closed, or both are all wrong.

Your Answer : (Not Answered)

Correct Answer : B

I is TRUE and II is FALSE.

I: for \(x \in C_{00}\), \(|f(x)| \le \sum|x_{n}|/\sqrt{n} \le \|x\|_{1}\), so \(f\) is bounded in the \(\ell^{1}\) norm. Since \(C_{00}\) is dense in \(\ell^{1}\), \(f\) extends to a continuous linear functional \(F\) on \(\ell^{1}\) (given by the bounded sequence \(1/\sqrt{n}\)).

II: a continuous extension to \(\ell^{2}\) would be \(G(x) = \sum x_{n}c_{n}\) with \((c_{n}) \in \ell^{2}\), forcing \(c_{n} = 1/\sqrt{n}\). But \(\sum 1/n\) diverges, so \((1/\sqrt{n}) \notin \ell^{2}\). Concretely, \(x = (1/\sqrt{n})_{n \le N}\) has \(\|x\|_{2}^{2} = f(x) = \sum_{n \le N}1/n\), so \(f(x)/\|x\|_{2}\) is unbounded and \(f\) is not continuous on \(\ell^{2}\).

Your Answer : (Not Answered)

Correct Answer : D

NEITHER S NOR T is a compact linear map.

A compact operator maps the bounded sequence of unit vectors \(e_{n}\) to a sequence with a convergent subsequence.

S is the right shift: \(Se_{n} = e_{n+1}\), and \(\|e_{n+1} - e_{m+1}\| = \sqrt{2}\) for \(n \ne m\), so no subsequence converges. S is an isometry on an infinite-dimensional space, so it is not compact.

T is the projection onto the even coordinates: \(Te_{2n} = e_{2n}\), again a sequence with mutual distances \(\sqrt{2}\). T is the identity on an infinite-dimensional subspace, so it is not compact either.

Your Answer : (Not Answered)

Correct Answer : A

The incorrect statement is \(S_{1} \subseteq S_{2} \Rightarrow S_{1}^{\perp} \subseteq S_{2}^{\perp}\).

Taking orthogonal complements reverses inclusion. A vector orthogonal to every element of the larger set \(S_{2}\) is orthogonal to every element of \(S_{1}\), so the true statement is \(S_{2}^{\perp} \subseteq S_{1}^{\perp}\). For example, \(\{0\} \subseteq H\) but \(\{0\}^{\perp} = H\) is not contained in \(H^{\perp} = \{0\}\).

The other three are correct. Every \(s \in S\) is orthogonal to every element of \(S^{\perp}\), so \(S \subseteq (S^{\perp})^{\perp}\). Every vector is orthogonal to 0, so \(\{0\}^{\perp} = H\). And \(S^{\perp}\) is an intersection of kernels of the continuous functionals \(x \mapsto \langle x, s \rangle\), hence always closed.

Your Answer : (Not Answered)

Correct Answer : D

Both \(f(X)\) and \(X\) are uncountable.

A continuous image of a connected space is connected, and the connected subsets of \(\mathbb{R}\) are intervals. Since \(f\) is not constant, \(f(X)\) contains two different values and hence the whole interval between them, which is uncountable.

\(f\) maps \(X\) onto \(f(X)\), so \(X\) has at least as many points as \(f(X)\) and is uncountable too.

This rules out every option in which \(f(X)\) or \(X\) is countable.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct 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.