Question

Consider \(\mathbb{R}\) with the topology
\[ \tau = \{A \subseteq \mathbb{R} : A^c \text{ is finite}\} \cup \{\phi\}. \]
Which of the following statements is/are TRUE?

• A. \((\mathbb{R}, \tau)\) is a Hausdorff space.
• B. Any finite subset of \((\mathbb{R}, \tau)\) is closed.
• C. \((\mathbb{R}, \tau)\) is compact.
• D. \((\mathbb{R}, \tau)\) is connected.

Hint:

This is the cofinite topology: every nonempty open set has a finite complement. Think about whether two such sets can ever be disjoint.

Answer 1

The Correct Option is B, C, D

The topology $\tau = \{A \subseteq \mathbb{R} : A^c \text{ is finite}\} \cup \{\phi\}$ is the well known cofinite topology. Its single defining feature, that every nonempty open set misses only finitely many points of $\mathbb{R}$, drives every property below.

1. (A) Hausdorff: False. If two nonempty open sets $U$ and $V$ were disjoint, each would have to sit inside the complement of the other. Since complements are finite, both $U$ and $V$ would have to be finite. But a nonempty open set already has a finite complement in an infinite space $\mathbb{R}$, so it cannot itself be finite as well (that would make $\mathbb{R}$ finite). So no two nonempty open sets can ever be disjoint, and the Hausdorff separation property fails completely.
2. (B) Finite subsets are closed: True. Take any finite set $F$. Its complement $F^c$ has complement equal to $F$ itself, which is finite by assumption, so $F^c$ satisfies the exact rule that makes a set open in $\tau$. Hence $F^c$ is open, which by definition makes $F$ closed.
3. (C) Compact: True. Given any open cover of $\mathbb{R}$, choose one nonempty set $U$ from it. Only finitely many points of $\mathbb{R}$ lie outside $U$, and each of those points is covered by at least one more set from the cover. Adding those finitely many sets to $U$ produces a finite subcover, so every open cover reduces to a finite one.
4. (D) Connected: True. A space is disconnected only if it splits into two disjoint nonempty open sets. Here, if $\mathbb{R} = U \sqcup V$ with both open and nonempty, then $V$ sits inside the finite set $U^c$, so $V$ is finite; but $V$ being open and nonempty also forces $V^c$ finite, and then $\mathbb{R} = V \cup V^c$ would be a union of two finite sets, which is impossible since $\mathbb{R}$ is infinite. So no such split can exist.

Every open set except the empty set is almost all of $\mathbb{R}$, missing only a finite chunk. That single idea is what breaks Hausdorff separation, yet guarantees closedness of finite sets, compactness, and connectedness all at once.

Let's summarize:

• Two nonempty open sets in this topology can never be disjoint, so Hausdorff fails.
• Finiteness of complements is exactly what makes finite sets closed, covers reducible to finite subcovers, and the whole space impossible to split into two open pieces.

The correct options are (B), (C), and (D).

\[ \boxed{\text{(B), (C) and (D)}} \]