Topology Practice Questions

\(\{b,c\} \notin \tau\), so it is not open. Its complement is \(\{a\} \in \tau\), which is open, so \(\{b,c\}\) is closed. Hence it is closed but not open.

Answer: B)

In the discrete topology every subset is open. Since the complement of any set is also a subset, every complement is open, so every subset is also closed.

Answer: C)

In the finite complement topology, a set is open iff its complement is finite (or the set is empty). \(\mathbb{R} \setminus \{1,2,3\}\) has complement \(\{1,2,3\}\), which is finite.

Answer: B)

\((0,1) = \bigcup_{n=1}^{\infty}[1/n, 1)\), which is a union of basis elements, so it is open in \(\mathbb{R}_\ell\). The others are not open: \((0,1]\) is not a union of half-open intervals of the form \([a,b)\); \([0,1]\) and \(\{0\}\) also fail.

Answer: A)

\([0,1)\) is a basis element so it is open. Its complement is \((-\infty, 0) \cup [1, \infty)\). Now \((-\infty,0) = \bigcup_{n}[-n, 0)\) and \([1,\infty) = \bigcup_{n}[1, n)\), both open. So the complement is open, meaning \([0,1)\) is also closed. It is clopen.

Answer: B)

In \(\mathbb{R}_\ell\), the only connected subsets are singletons and the empty set. Any set with more than one point can be separated: if \(a < b\) are in a set \(S\), then \(S = (S \cap (-\infty, b)) \cup (S \cap [b, \infty))\) is a separation since both parts are open in the subspace topology.

Answer: C)

By the Heine–Borel theorem, a subset of \(\mathbb{R}\) is compact iff it is closed and bounded. \([0,1]\) is both closed and bounded. \((0,1)\) is not closed, \([0,\infty)\) is not bounded, \(\mathbb{Q}\cap[0,1]\) is not closed, and \(\mathbb{Z}\) is not bounded.

Answer: D)

In the discrete topology every singleton \(\{x\}\) is open. So for any infinite set \(S\), the cover \(\{\{x\} : x \in S\}\) has no finite subcover. Thus only finite sets are compact.

Answer: C)

In the discrete topology, every subset is both open and closed. So if \(S\) has two distinct points \(a, b\), then \(\{a\}\) and \(S\setminus\{a\}\) form a separation. Only singletons and \(\emptyset\) are connected.

Answer: C)

The lower limit topology is strictly finer than the standard topology. A continuous map requires preimages of open sets to be open. The set \([0,1)\) is open in \(\mathbb{R}_\ell\), but \(f^{-1}([0,1)) = (-1, 1)\) is open in the standard topology. However, \([4, 9)\) is open in \(\mathbb{R}_\ell\) and \(f^{-1}([4,9)) = (-3,-2] \cup [2,3)\), and \((-3,-2]\) is not open in the standard topology. So \(f\) is not continuous.

Answer: B)

If \(X\) has the discrete topology, every subset of \(X\) is open. So the preimage of any open set in \(Y\) is automatically open in \(X\). Therefore every function from a discrete space is continuous.

Answer: C)

The only open sets in \(Y\) are \(\emptyset\) and \(Y\). For any function \(f\), \(f^{-1}(\emptyset) = \emptyset\) and \(f^{-1}(Y) = X\), both of which are open in any topology on \(X\). So every function into the indiscrete topology is continuous.

Answer: B)

Every real number is a limit point of \(\mathbb{Q}\) since every open interval contains a rational number (density of \(\mathbb{Q}\)). Thus \(\overline{\mathbb{Q}} = \mathbb{R}\).

Answer: B)

A point \(x\) is a limit point of \(S\) if every open set containing \(x\) contains a point of \(S\) different from \(x\). The sequence \(1/n \to 0\), so \(0\) is a limit point. Each \(1/n\) is isolated (can be separated from other points of \(S\) by a small interval), so no element of \(S\) is a limit point of \(S\).

Answer: B)

The boundary is \(\overline{S} \setminus \text{int}(S)\). Here \(\text{int}(S) = (0,1) \cup (1,2)\) (it's already open) and \(\overline{S} = [0,2]\). So the boundary is \([0,2] \setminus ((0,1)\cup(1,2)) = \{0, 1, 2\}\).

Answer: B)

In the finite complement topology on an infinite set, any two nonempty open sets must intersect (their complements are finite, so the intersection's complement is also finite, hence the intersection is nonempty). Since we cannot find disjoint open neighborhoods of distinct points, it is not Hausdorff.

Answer: D)

A subset of \(\mathbb{R}\) is connected if and only if it is an interval. This includes all forms: open, closed, half-open, rays, singletons, \(\emptyset\), and \(\mathbb{R}\) itself. (In \(\mathbb{R}\), intervals and convex sets are the same, but the standard terminology is intervals.)

Answer: C)

\((0,1)\) and \(\mathbb{R}\) are homeomorphic via \(f(x) = \tan(\pi x - \pi/2)\). The other pairs fail: \((0,1)\) is not compact but \([0,1]\) is; \([0,1]\) is compact but \(\mathbb{R}\) is not; \(S^1\) is compact but removing a point disconnects neither while removing an endpoint of \([0,1]\) does not disconnect it either — but \(S^1\) has no "endpoint" to remove, removing any point of \(S^1\) leaves it connected, while removing \(0\) from \([0,1]\) leaves \((0,1]\), which is connected, but removing an interior point from \([0,1]\) disconnects it while removing any point from \(S^1\) does not; \([0,1]\) and \([0,1)\) differ in compactness.

Answer: B)

Continuous functions preserve connectedness: the continuous image of a connected set is connected. They do not necessarily preserve openness, closedness, or compactness (compactness is preserved, but that's not the question about \(A\) being connected).

Answer: C)

The continuous image of a compact set is compact. This is a fundamental theorem: if \(\{U_\alpha\}\) covers \(f(K)\), then \(\{f^{-1}(U_\alpha)\}\) covers \(K\), extract a finite subcover, and the corresponding \(U_\alpha\)'s cover \(f(K)\).

Answer: B)

The topologist's sine curve is the classic example of a connected space that is not path-connected. The curve \(\{(x, \sin(1/x)): x > 0\}\) is path-connected and its closure includes the segment \(\{0\}\times[-1,1]\), making the whole set connected. But there is no continuous path from any point on the vertical segment to the oscillating curve.

Answer: B)

Compactness is an intrinsic property — a subspace is compact iff every open cover (in the subspace topology) has a finite subcover. By Heine–Borel, a subset of \(\mathbb{R}\) is compact iff it is closed and bounded. \((0,1)\) is bounded but not closed, so it is not compact. Equivalently, the cover \(\{(1/n, 1) : n \geq 1\}\) has no finite subcover.

Answer: C)

Yes. Let \(\{U_\alpha\}\) be any open cover of \(\mathbb{R}\). Pick one \(U_{\alpha_0}\). Its complement \(\mathbb{R}\setminus U_{\alpha_0}\) is finite, say \(\{x_1, \ldots, x_n\}\). For each \(x_i\), pick \(U_{\alpha_i}\) containing \(x_i\). Then \(\{U_{\alpha_0}, U_{\alpha_1}, \ldots, U_{\alpha_n}\}\) is a finite subcover. So \(\mathbb{R}\) is compact in this topology.

Answer: A)

In the finite complement topology on an infinite set, any two nonempty open sets intersect (since each has finite complement, their intersection has complement that is a union of two finite sets, hence finite, so the intersection is nonempty). Therefore there is no separation of \(\mathbb{R}\) into two disjoint nonempty open sets, so \(\mathbb{R}\) is connected.

Answer: B)

Consider the open set \( (-1, 1) \). Then \( f((-1,1)) = [0, 1) \), which is not open in \(\mathbb{R}\). Since the image of an open set is not open, \(f\) is not an open map.

Answer: B)

Boundedness is a metric property, not a topological one. For example, \((0,1)\) and \(\mathbb{R}\) are homeomorphic, but \((0,1)\) is bounded while \(\mathbb{R}\) is not. Compactness, connectedness, Hausdorff, and openness are all topological properties preserved by homeomorphisms.

Answer: D)

In the K-topology, the set \((-1, 2) \setminus K\) is open (it is a basis element). This means every point not in \(K\) near \(K\) has an open neighborhood missing \(K\). More precisely, \(\mathbb{R}\setminus K\) is open since for any \(x \notin K\): if \(x \notin (0,1]\), use a standard open interval avoiding \(K\); if \(x \in (0,1]\setminus K\), use a basis element of the form \((a,b)\setminus K\). So \(K\) is closed.

Answer: B)

\(\mathbb{R}_K\) is not connected. The sets \((-\infty, 0) \cup (0,\infty) \setminus K\) and ... more directly: \((-\infty, 0)\) is open and \([0,\infty)\) is open in the K-topology? No. The correct separation is: \((-\infty, 0)\) is open (standard open set) and \((0, \infty) \setminus K\) together... Actually the key fact is that \(0\) cannot be separated from negative numbers. The correct argument: \(\mathbb{R}_K\) is not connected because \((-\infty, 0) \cup K^c\) provides a separation. Specifically, \((-\infty, 0)\) and \([0, \infty) \setminus K\) ... The simplest argument: consider \(U = (-1, 1) \setminus K\) which is open and contains \(0\), and \(K\) is closed. So \(\mathbb{R}\) can be separated as \(\mathbb{R} \setminus K\) (open) and ... This needs care. The answer is that \(\mathbb{R}_K\) is NOT connected.

Answer: B)

By the axioms of a topology, \(\emptyset\) and \(X\) are always open. Their complements (\(X\) and \(\emptyset\) respectively) are also open, so both are also closed. Hence \(\emptyset\) and \(X\) are always clopen in any topological space.

Answer: C)

In \(\mathbb{R}_\ell\), the set \([0,1)\) is clopen (both open and closed). Since \((0,1) \subset [0,1)\) and \([0,1)\) is closed, the closure of \((0,1)\) is contained in \([0,1)\). Is \(0\) a limit point of \((0,1)\)? Every open set containing \(0\) in \(\mathbb{R}_\ell\) contains a basis element \([0, \epsilon)\), which intersects \((0,1)\). So \(0\) is a limit point. Is \(1\) a limit point? The basis element \([1, 2)\) contains \(1\) but does not intersect \((0,1)\). So \(1\) is not a limit point. Thus the closure is \([0, 1)\).

Answer: B)

The fundamental group of the circle \( S^1 \) is isomorphic to the group of integers, \( \mathbb{Z} \). This can be shown using the winding number of a loop on the circle.

Answer: C) \( \mathbb{Z} \)