Connectedness and Compactness Questions and Answers

Q: Consider the following statements:P: The family of subsets \(\left\{A_{n} = \left(-\frac{1}{n}, \frac{1}{n}\right), n = 1, 2, \ldots\right\}\) satisfies the finite intersection property.Q: On an infinite set \(X\), a metric \(d : X \times X \to R\) is defined as \(d(x, y) = \begin{cases} 0, & x = y \\ 1, & x \neq y \end{cases}\). The metric space \((X, d)\) is compact.R: In a Frechet (\(T_{1}\)) topological space, every finite set is closed.S: If \(f : R \to X\) is continuous, where \(R\) is given the usual topology and \((X, \tau)\) is a Hausdorff (\(T_{2}\)) space, then \(f\) is a one-one function.Which of the above statements are correct?

The correct statements are P and R.P is true: the intervals \((-1/n, 1/n)\) are nested and all contain 0, so every finite subfamily has non-empty intersection. R is true: in a \(T_{1}\) space every singleton is closed, and a finite union of closed sets is closed.Q is false: with the discrete metric every singleton is an open ball of radius \(1/2\), and these open sets cover the infinite set \(X\) with no finite subcover, so \((X, d)\) is not compact. S is false: a constant map from \(R\) into any Hausdorff space is continuous but not one-one. Any option containing Q or S is therefore wrong.

Q: Let \(X\) be a connected topological space such that there exists a non-constant continuous function \(f : X \to \mathbb{R}\), where \(\mathbb{R}\) is equipped with the usual topology. Let \(f(X) = \{f(x) : x \in X\}\). Then

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.

Q: Let \(S = \{(x, \sin\frac{1}{x}) : 0 \lt x \le 1\}\) and \(T = S \cup \{(0,0)\}\). Under the usual metric on \(\mathbb{R}^{2}\),

Neither \(S\) nor \(T\) is closed.As \(x \to 0^{+}\), \(\sin\frac{1}{x}\) takes every value in \([-1, 1]\) infinitely often. So every point \((0, b)\) with \(-1 \le b \le 1\) is a limit point of \(S\), for instance \((0, 1)\) as the limit of the points with \(x = \frac{1}{\pi/2 + 2k\pi}\).None of these points lies in \(S\), so \(S\) is not closed. \(T\) adds only the origin and still misses \((0, 1)\), so \(T\) is not closed either. Its closure is \(S\) together with the whole segment \(\{0\} \times [-1, 1]\).

Q: Consider \(([0,1], T_{1})\), where \(T_{1}\) is the subspace topology induced by the Euclidean topology on \(\mathbb{R}\), and let \(T_{2}\) be any topology on [0, 1]. Consider the following statements:P : If \(T_{1}\) is a proper subset of \(T_{2}\), then \(([0,1], T_{2})\) is not compact.Q : If \(T_{2}\) is a proper subset of \(T_{1}\), then \(([0,1], T_{2})\) is not Hausdorff.Then

The answer is Both P and Q are TRUE.Both follow from one fact: a continuous bijection from a compact space onto a Hausdorff space is a homeomorphism. \(([0,1], T_{1})\) is compact and Hausdorff.P: if \(T_{2}\) strictly contains \(T_{1}\) and \(([0,1], T_{2})\) were compact, the identity from \(T_{2}\) to \(T_{1}\) would be a continuous bijection from a compact space to a Hausdorff one, hence a homeomorphism, forcing \(T_{2} = T_{1}\). So \(T_{2}\) is not compact.Q: if \(T_{2}\) is strictly contained in \(T_{1}\) and were Hausdorff, the identity from \(T_{1}\) to \(T_{2}\) would be a homeomorphism for the same reason, forcing equality. So \(T_{2}\) is not Hausdorff.A compact Hausdorff topology is both maximal compact and minimal Hausdorff, so every option calling P or Q false is wrong.

Q: Let X and Y be two topological spaces. A continuous map \(f : X \to Y\) is said to be proper if \(f^{-1}(K)\) is compact in X for every compact subset K of Y, where \(f^{-1}(K)\) is defined by \(f^{-1}(K) = \{x \in X : f(x) \in K\}\).Consider \(\mathbb{R}\) with the usual topology. If \(\mathbb{R} \setminus \{0\}\) has the subspace topology induced from \(\mathbb{R}\) and \(\mathbb{R} \times \mathbb{R}\) has the product topology, then which of the following maps is proper?

The proper map is \(f(x,y) = (x+y, y)\) on \(\mathbb{R} \times \mathbb{R}\).This f is linear with inverse \((u, v) \mapsto (u - v, v)\), so it is a homeomorphism of the plane. The preimage of a compact set is its image under the continuous inverse, which is compact. So f is proper.The others fail. The inclusion of \(\mathbb{R} \setminus \{0\}\) into \(\mathbb{R}\) pulls back [-1, 1] to \([-1,0) \cup (0,1]\), not compact. The projection (x, y) → x pulls back the point {0} to the whole y-axis, unbounded. And \(x^{2}-y^{2}\) pulls back {0} to the two lines y = ±x, also unbounded.

Q: Let\(A = \{(x, y) \in \mathbf{R}^{2}/x = 0, -1 \le y \le 1\}\)\[ \begin{aligned} B = \{&(x, y) \in \mathbf{R}^{2}/ \\ &0 \lt x \le \tfrac{1}{\pi}, y = \sin\tfrac{1}{x}\} \end{aligned} \]and \(X = A \cup B\). Then :

X is connected but not path connected.X is the topologist's sine curve: the graph B of \(\sin(1/x)\) on \((0, 1/\pi]\) together with the segment A on the \(y\)-axis from \(-1\) to 1. B is connected (a continuous image of an interval), and every point of A is a limit of points of B, because \(\sin(1/x)\) takes every value in \([-1, 1]\) on every interval \((0, \epsilon)\). So \(X = A \cup B\) lies between B and its closure and is connected.No path joins a point of A to a point of B: a path would have to reach \(x = 0\) while its \(y\)-coordinate oscillates between \(-1\) and 1 infinitely often, which contradicts continuity. So X is not path connected, ruling out A, C and D.

Q: Which of the following subsets of \(\mathbf{R}^{2}\) are not compact ?(i) a circle(ii) a parabola(iii) \(S = \left\{(x, y) \in \mathbf{R}^{2}/y = \sin\dfrac{1}{x}, 0 \lt x \le 1\right\}\)(iv) \(T = \{(x, y) \in \mathbf{R}^{2}/|x| + |y| = 4\}\)

The non-compact sets are (ii) and (iii).In \(\mathbf{R}^{2}\), compact means closed and bounded (Heine-Borel). A parabola is closed but unbounded, so not compact. S, the graph of \(\sin(1/x)\) on \((0, 1]\), is bounded but not closed: its closure adds the segment from \((0, -1)\) to \((0, 1)\), which S does not contain.A circle is closed and bounded, so compact. T, \(|x| + |y| = 4\), is the boundary of a square (a diamond) through \((\pm 4, 0)\) and \((0, \pm 4)\): closed and bounded, so also compact. That rules out every option containing (i) or (iv).

Q: Let X be a locally connected topological space. Then which of the following is true ?

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.

Q: Consider the following statements :(a) There is a continuous map from [0, 1] onto (0, 1)(b) There is a continuous map from (0, 1) onto [0, 1]Then :

The answer is only (b) is true.(a) is false: [0, 1] is compact and the continuous image of a compact set is compact, but (0, 1) is not compact. (Equivalently, a continuous function on [0, 1] attains its maximum and minimum, so its image is a closed interval.)(b) is true: for example \(f(x) = \tfrac{1}{2}(1 + \sin(4\pi x))\) is continuous on (0, 1) and takes the values 0 and 1 inside the interval, at \(x = 3/8\) and \(x = 1/8\), so its image is all of [0, 1]. Compactness is not needed in the domain to reach a compact image.

Q: Let \(X = \{(x, y) \in \mathbf{R}^{2} : |x|\ |y| = 1\}\) and \(Y = \{(x, y) \in \mathbf{R}^{2} : |x| + |y| = 1\}\). Then :

The answer is Y is compact, but X is not compact.By Heine-Borel, a subset of \(\mathbf{R}^{2}\) is compact exactly when it is closed and bounded. Both sets are closed, being level sets of continuous functions.Y, defined by \(|x| + |y| = 1\), is the square with vertices \((\pm 1, 0)\) and \((0, \pm 1)\); it is bounded, so it is compact. X, defined by \(|x|\,|y| = 1\), is the four branches of the hyperbolas \(xy = \pm 1\); it contains \((n, 1/n)\) for every \(n\), so it is unbounded and not compact.

Consider the following statements:

P: The family of subsets \(\left\{A_{n} = \left(-\frac{1}{n}, \frac{1}{n}\right), n = 1, 2, \ldots\right\}\) satisfies the finite intersection property.
Q: On an infinite set \(X\), a metric \(d : X \times X \to R\) is defined as \(d(x, y) = \begin{cases} 0, & x = y \\ 1, & x \neq y \end{cases}\). The metric space \((X, d)\) is compact.
R: In a Frechet (\(T_{1}\)) topological space, every finite set is closed.
S: If \(f : R \to X\) is continuous, where \(R\) is given the usual topology and \((X, \tau)\) is a Hausdorff (\(T_{2}\)) space, then \(f\) is a one-one function.

Which of the above statements are correct?

A.P and R
B.P and S
C.R and S
D.Q and S

Exam Year(s)

GATE Mathematics - 2012,

Answer: A

The correct statements are P and R.

P is true: the intervals \((-1/n, 1/n)\) are nested and all contain 0, so every finite subfamily has non-empty intersection. R is true: in a \(T_{1}\) space every singleton is closed, and a finite union of closed sets is closed.

Q is false: with the discrete metric every singleton is an open ball of radius \(1/2\), and these open sets cover the infinite set \(X\) with no finite subcover, so \((X, d)\) is not compact. S is false: a constant map from \(R\) into any Hausdorff space is continuous but not one-one. Any option containing Q or S is therefore wrong.

Discuss in Forum

Let \(X\) be a connected topological space such that there exists a non-constant continuous function \(f : X \to \mathbb{R}\), where \(\mathbb{R}\) is equipped with the usual topology. Let \(f(X) = \{f(x) : x \in X\}\). Then

A.\(X\) is countable but \(f(X)\) is uncountable
B.\(f(X)\) is countable but \(X\) is uncountable
C.both \(f(X)\) and \(X\) are countable
D.both \(f(X)\) and \(X\) are uncountable

Exam Year(s)

GATE Mathematics - 2015,

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.

Discuss in Forum

Let \(S = \{(x, \sin\frac{1}{x}) : 0 \lt x \le 1\}\) and \(T = S \cup \{(0,0)\}\). Under the usual metric on \(\mathbb{R}^{2}\),

A.\(S\) is closed but \(T\) is NOT closed
B.\(T\) is closed but \(S\) is NOT closed
C.both \(S\) and \(T\) are closed
D.neither \(S\) nor \(T\) is closed

Exam Year(s)

GATE Mathematics - 2015,

Answer: D

Neither \(S\) nor \(T\) is closed.

As \(x \to 0^{+}\), \(\sin\frac{1}{x}\) takes every value in \([-1, 1]\) infinitely often. So every point \((0, b)\) with \(-1 \le b \le 1\) is a limit point of \(S\), for instance \((0, 1)\) as the limit of the points with \(x = \frac{1}{\pi/2 + 2k\pi}\).

None of these points lies in \(S\), so \(S\) is not closed. \(T\) adds only the origin and still misses \((0, 1)\), so \(T\) is not closed either. Its closure is \(S\) together with the whole segment \(\{0\} \times [-1, 1]\).

Discuss in Forum

Consider \(([0,1], T_{1})\), where \(T_{1}\) is the subspace topology induced by the Euclidean topology on \(\mathbb{R}\), and let \(T_{2}\) be any topology on [0, 1]. Consider the following statements:

P : If \(T_{1}\) is a proper subset of \(T_{2}\), then \(([0,1], T_{2})\) is not compact.
Q : If \(T_{2}\) is a proper subset of \(T_{1}\), then \(([0,1], T_{2})\) is not Hausdorff.

Then

A.P is TRUE and Q is FALSE
B.Both P and Q are TRUE
C.Both P and Q are FALSE
D.P is FALSE and Q is TRUE

Exam Year(s)

GATE Mathematics - 2022,

Answer: B

The answer is Both P and Q are TRUE.

Both follow from one fact: a continuous bijection from a compact space onto a Hausdorff space is a homeomorphism. \(([0,1], T_{1})\) is compact and Hausdorff.

P: if \(T_{2}\) strictly contains \(T_{1}\) and \(([0,1], T_{2})\) were compact, the identity from \(T_{2}\) to \(T_{1}\) would be a continuous bijection from a compact space to a Hausdorff one, hence a homeomorphism, forcing \(T_{2} = T_{1}\). So \(T_{2}\) is not compact.

Q: if \(T_{2}\) is strictly contained in \(T_{1}\) and were Hausdorff, the identity from \(T_{1}\) to \(T_{2}\) would be a homeomorphism for the same reason, forcing equality. So \(T_{2}\) is not Hausdorff.

A compact Hausdorff topology is both maximal compact and minimal Hausdorff, so every option calling P or Q false is wrong.

Discuss in Forum

Let X and Y be two topological spaces. A continuous map \(f : X \to Y\) is said to be proper if \(f^{-1}(K)\) is compact in X for every compact subset K of Y, where \(f^{-1}(K)\) is defined by \(f^{-1}(K) = \{x \in X : f(x) \in K\}\).

Consider \(\mathbb{R}\) with the usual topology. If \(\mathbb{R} \setminus \{0\}\) has the subspace topology induced from \(\mathbb{R}\) and \(\mathbb{R} \times \mathbb{R}\) has the product topology, then which of the following maps is proper?

A.\(f : \mathbb{R} \setminus \{0\} \to \mathbb{R}\) defined by \(f(x) = x\)
B.\(f : \mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) defined by \(f(x,y) = (x+y, y)\)
C.\(f : \mathbb{R} \times \mathbb{R} \to \mathbb{R}\) defined by \(f(x,y) = x\)
D.\(f : \mathbb{R} \times \mathbb{R} \to \mathbb{R}\) defined by \(f(x,y) = x^{2}-y^{2}\)

Exam Year(s)

GATE Mathematics - 2023,

Answer: B

The proper map is \(f(x,y) = (x+y, y)\) on \(\mathbb{R} \times \mathbb{R}\).

This f is linear with inverse \((u, v) \mapsto (u - v, v)\), so it is a homeomorphism of the plane. The preimage of a compact set is its image under the continuous inverse, which is compact. So f is proper.

The others fail. The inclusion of \(\mathbb{R} \setminus \{0\}\) into \(\mathbb{R}\) pulls back [-1, 1] to \([-1,0) \cup (0,1]\), not compact. The projection (x, y) → x pulls back the point {0} to the whole y-axis, unbounded. And \(x^{2}-y^{2}\) pulls back {0} to the two lines y = ±x, also unbounded.

Discuss in Forum

Let

\(A = \{(x, y) \in \mathbf{R}^{2}/x = 0, -1 \le y \le 1\}\)

\[ \begin{aligned} B = \{&(x, y) \in \mathbf{R}^{2}/ \\ &0 \lt x \le \tfrac{1}{\pi}, y = \sin\tfrac{1}{x}\} \end{aligned} \]

and \(X = A \cup B\). Then :

A.X is connected and path connected
B.X is connected but not path connected
C.X is not connected but path connected
D.X is neither connected nor path connected

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

X is connected but not path connected.

X is the topologist's sine curve: the graph B of \(\sin(1/x)\) on \((0, 1/\pi]\) together with the segment A on the \(y\)-axis from \(-1\) to 1. B is connected (a continuous image of an interval), and every point of A is a limit of points of B, because \(\sin(1/x)\) takes every value in \([-1, 1]\) on every interval \((0, \epsilon)\). So \(X = A \cup B\) lies between B and its closure and is connected.

No path joins a point of A to a point of B: a path would have to reach \(x = 0\) while its \(y\)-coordinate oscillates between \(-1\) and 1 infinitely often, which contradicts continuity. So X is not path connected, ruling out A, C and D.

Discuss in Forum

Which of the following subsets of \(\mathbf{R}^{2}\) are not compact ?

(i) a circle
(ii) a parabola
(iii) \(S = \left\{(x, y) \in \mathbf{R}^{2}/y = \sin\dfrac{1}{x}, 0 \lt x \le 1\right\}\)
(iv) \(T = \{(x, y) \in \mathbf{R}^{2}/|x| + |y| = 4\}\)

A.(i) and (ii)
B.(ii) and (iii)
C.(iii) and (iv)
D.(ii), (iii) and (iv)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The non-compact sets are (ii) and (iii).

In \(\mathbf{R}^{2}\), compact means closed and bounded (Heine-Borel). A parabola is closed but unbounded, so not compact. S, the graph of \(\sin(1/x)\) on \((0, 1]\), is bounded but not closed: its closure adds the segment from \((0, -1)\) to \((0, 1)\), which S does not contain.

A circle is closed and bounded, so compact. T, \(|x| + |y| = 4\), is the boundary of a square (a diamond) through \((\pm 4, 0)\) and \((0, \pm 4)\): closed and bounded, so also compact. That rules out every option containing (i) or (iv).

Discuss in Forum

Let X be a locally connected topological space. Then which of the following is true ?

A.X is connected
B.X is locally path connected
C.Every component of an open set of X is open in X
D.Every component of X is also a path component of X

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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.

Discuss in Forum

Consider the following statements :

(a) There is a continuous map from [0, 1] onto (0, 1)
(b) There is a continuous map from (0, 1) onto [0, 1]

Then :

A.only (a) is true
B.only (b) is true
C.both (a) and (b) are true
D.both (a) and (b) are false

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The answer is only (b) is true.

(a) is false: [0, 1] is compact and the continuous image of a compact set is compact, but (0, 1) is not compact. (Equivalently, a continuous function on [0, 1] attains its maximum and minimum, so its image is a closed interval.)

(b) is true: for example \(f(x) = \tfrac{1}{2}(1 + \sin(4\pi x))\) is continuous on (0, 1) and takes the values 0 and 1 inside the interval, at \(x = 3/8\) and \(x = 1/8\), so its image is all of [0, 1]. Compactness is not needed in the domain to reach a compact image.

Discuss in Forum

10.

Let \(X = \{(x, y) \in \mathbf{R}^{2} : |x|\ |y| = 1\}\) and \(Y = \{(x, y) \in \mathbf{R}^{2} : |x| + |y| = 1\}\). Then :

A.X and Y are compact
B.X is compact, but Y is not compact
C.Y is compact, but X is not compact
D.neither X nor Y is compact

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The answer is Y is compact, but X is not compact.

By Heine-Borel, a subset of \(\mathbf{R}^{2}\) is compact exactly when it is closed and bounded. Both sets are closed, being level sets of continuous functions.

Y, defined by \(|x| + |y| = 1\), is the square with vertices \((\pm 1, 0)\) and \((0, \pm 1)\); it is bounded, so it is compact. X, defined by \(|x|\,|y| = 1\), is the four branches of the hyperbolas \(xy = \pm 1\); it contains \((n, 1/n)\) for every \(n\), so it is unbounded and not compact.

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