Separation and Countability Axioms Questions and Answers

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.

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.

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).

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 \(\mathbf{R}\).

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.

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.