Separation and Countability Axioms Questions and Answers

Q: Let \(f : X \to Y\) be a continuous map from a Hausdorff topological space \(X\) to a metric space \(Y\). Consider the following two statements:P: \(f\) is a closed map and the inverse image \(f^{-1}(y) = \{x \in X : f(x) = y\}\) is compact for each \(y \in Y\).Q: For every compact subset \(K \subset Y\), the inverse image \(f^{-1}(K)\) is a compact subset of \(X\).Which one of the following is true?

P and Q are equivalent.P implies Q: a closed continuous map with compact fibres (a perfect map) pulls compact sets back to compact sets. This is a standard theorem and needs no condition on \(Y\) beyond what is given.Q implies P: the fibres \(f^{-1}(y)\) are compact because \(\{y\}\) is compact. For closedness, let \(C\) be closed in \(X\) and \(f(c_{n}) \to y\) with \(c_{n} \in C\); sequences suffice because \(Y\) is a metric space. The set \(K = \{f(c_{n})\} \cup \{y\}\) is compact, so the \(c_{n}\) lie in the compact set \(f^{-1}(K) \cap C\) and have a cluster point \(c \in C\). By continuity \(f(c)\) is a cluster point of \(f(c_{n})\), so \(f(c) = y\) and \(y \in f(C)\). Hence \(f\) is closed.Since each statement implies the other, the options asserting a one-way implication or none are wrong.

Q: Consider the following statements.I. Every compact Hausdorff space is normal.II. Every metric space is normal.Which one of the following is correct?

Both I and II are TRUE.I: in a compact Hausdorff space, a point and a disjoint compact (closed) set can be separated, and repeating the argument with finite subcovers separates two disjoint closed sets. So the space is normal.II: in a metric space, disjoint closed sets \(A\) and \(B\) are separated by the open sets \(\{x : d(x, A) \lt d(x, B)\}\) and \(\{x : d(x, B) \lt d(x, A)\}\). So every metric space is normal. Options declaring either statement false are wrong.

Q: Which of the following spaces need not be Hausdorff ?

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).

Q: Let X be a second countable space and \(f : X \to Y\) be continuous open surjective map. Then :

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.

Q: Let X be a separable first countable space, then X × X in the product topology is :

X × X is first countable and separable.First countable: if \(\{U_{n}\}\) and \(\{V_{m}\}\) are countable local bases at \(x\) and \(y\), then \(\{U_{n} \times V_{m}\}\) is a countable local base at \((x, y)\).Separable: if D is a countable dense subset of X, then D × D is countable and meets every basic open set \(U \times V\), since D meets both U and V; so it is dense in X × X.Both properties are preserved by finite (indeed countable) products, so the options saying either may fail are wrong.

Let \(f : X \to Y\) be a continuous map from a Hausdorff topological space \(X\) to a metric space \(Y\). Consider the following two statements:

P: \(f\) is a closed map and the inverse image \(f^{-1}(y) = \{x \in X : f(x) = y\}\) is compact for each \(y \in Y\).

Q: For every compact subset \(K \subset Y\), the inverse image \(f^{-1}(K)\) is a compact subset of \(X\).

Which one of the following is true?

A.Q implies P but P does NOT imply Q
B.P implies Q but Q does NOT imply P
C.P and Q are equivalent
D.neither P implies Q nor Q implies P

Exam Year(s)

GATE Mathematics - 2018,

Answer: C

P and Q are equivalent.

P implies Q: a closed continuous map with compact fibres (a perfect map) pulls compact sets back to compact sets. This is a standard theorem and needs no condition on \(Y\) beyond what is given.

Q implies P: the fibres \(f^{-1}(y)\) are compact because \(\{y\}\) is compact. For closedness, let \(C\) be closed in \(X\) and \(f(c_{n}) \to y\) with \(c_{n} \in C\); sequences suffice because \(Y\) is a metric space. The set \(K = \{f(c_{n})\} \cup \{y\}\) is compact, so the \(c_{n}\) lie in the compact set \(f^{-1}(K) \cap C\) and have a cluster point \(c \in C\). By continuity \(f(c)\) is a cluster point of \(f(c_{n})\), so \(f(c) = y\) and \(y \in f(C)\). Hence \(f\) is closed.

Since each statement implies the other, the options asserting a one-way implication or none are wrong.

Discuss in Forum

Consider the following statements.

I. Every compact Hausdorff space is normal.
II. Every metric space is normal.

Which one of the following is correct?

A.Both I and II are TRUE
B.I is TRUE and II is FALSE
C.I is FALSE and II is TRUE
D.Both I and II are FALSE

Exam Year(s)

GATE Mathematics - 2024,

Answer: A

Both I and II are TRUE.

I: in a compact Hausdorff space, a point and a disjoint compact (closed) set can be separated, and repeating the argument with finite subcovers separates two disjoint closed sets. So the space is normal.

II: in a metric space, disjoint closed sets \(A\) and \(B\) are separated by the open sets \(\{x : d(x, A) \lt d(x, B)\}\) and \(\{x : d(x, B) \lt d(x, A)\}\). So every metric space is normal. Options declaring either statement false are wrong.

Discuss in Forum

Which of the following spaces need not be Hausdorff ?

A.X is a topological space such that \(\Delta = \{x \times x/x \in X\}\) is closed in \(X \times X\)
B.X is a metric space
C.X is an Indiscrete Topological space
D.X is the product of two Hausdorff spaces

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

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).

Discuss in Forum

With the induced topology by the metric \(d(x, y) = |x - y|\),

A.\(\mathbf{R}\) and \(\mathbf{Q}\) are of second category
B.\(\mathbf{R}\) is of second category and \(\mathbf{Q}\) is not
C.\(\mathbf{Q}\) is of second category and \(\mathbf{R}\) is not
D.Both \(\mathbf{R}\) and \(\mathbf{Q}\) are not of second category

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

\(\mathbf{R}\) is of second category and \(\mathbf{Q}\) is not.

\(\mathbf{R}\) with the usual metric is complete, so by the Baire category theorem it is not a countable union of nowhere dense sets: it is of second category in itself.

\(\mathbf{Q}\) is countable, and each singleton \(\{q\}\) is nowhere dense in \(\mathbf{Q}\) (a single point has empty interior there). So \(\mathbf{Q} = \bigcup \{q\}\) is a countable union of nowhere dense sets, of first category. That rules out the options calling \(\mathbf{Q}\) second category or denying it to \(\mathbf{R}\).

Discuss in Forum

Let X be a second countable space and \(f : X \to Y\) be continuous open surjective map. Then :

A.Y is second countable
B.Y is separable, but not second countable
C.Y is first countable but not separable
D.Y is separable, but not first countable

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

Discuss in Forum

Let X be a separable first countable space, then X × X in the product topology is :

A.first countable and separable
B.first countable, but may not always be separable
C.separable and may not always be first countable
D.neither always first countable nor always separable

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: A

X × X is first countable and separable.

First countable: if \(\{U_{n}\}\) and \(\{V_{m}\}\) are countable local bases at \(x\) and \(y\), then \(\{U_{n} \times V_{m}\}\) is a countable local base at \((x, y)\).

Separable: if D is a countable dense subset of X, then D × D is countable and meets every basic open set \(U \times V\), since D meets both U and V; so it is dense in X × X.

Both properties are preserved by finite (indeed countable) products, so the options saying either may fail are wrong.

Separation and Countability Axioms Questions and Answers

Topics

Other Subjects

Tags: Separation and Countability Axioms Questions and Answers

About Us

Separation and Countability Axioms Questions and Answers

Topics

Other Subjects

Tags: Separation and Countability Axioms Questions and Answers

About Us

Stay informed about all important Dates & Notifications