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

Step 1: Recognize the topology.
This \(\tau\) is the standard cofinite (finite complement) topology on \(\mathbb{R}\): the open sets are exactly the empty set and every subset whose complement is finite. Since \(\mathbb{R}\) is infinite, this topology behaves very differently from the usual topology, and it is a classic example used to separate topological properties from each other.

Step 2: Check Hausdorff, option (A).
Take any two nonempty open sets \(U,V \in \tau\). Their complements \(U^c\) and \(V^c\) are both finite. If \(U\) and \(V\) were disjoint, then \(V \subseteq U^c\), which would force \(V\) to be finite. But \(V\) is a nonempty open set, so \(V^c\) is finite too, and a set with a finite complement inside an infinite space \(\mathbb{R}\) cannot itself be finite (a finite set union its finite complement can't cover all of \(\mathbb{R}\)). This contradiction shows any two nonempty open sets must intersect, so we can never separate two distinct points with disjoint open neighborhoods. \((\mathbb{R},\tau)\) is NOT Hausdorff. Option (A) is FALSE.

Step 3: Check that finite sets are closed, option (B).
A set \(F\) is closed exactly when its complement \(F^c\) is open. If \(F\) is finite, then \(F^c\) has complement \(F\), which is finite by assumption, so \(F^c \in \tau\) directly by the definition of \(\tau\). So every finite set is closed. Option (B) is TRUE.

Step 4: Check compactness, option (C).
Let \(\{U_i\}_{i \in I}\) be any open cover of \(\mathbb{R}\). Since the cover is nonempty and its union is all of \(\mathbb{R}\), pick one index \(i_0\) with \(U_{i_0}\) nonempty. Then \(U_{i_0}^c = \{p_1,\dots,p_n\}\) is a finite set of points not covered by \(U_{i_0}\). Each \(p_j\) belongs to \(\mathbb{R}\), so each \(p_j\) lies in some \(U_{i_j}\) from the cover. Then \(\{U_{i_0},U_{i_1},\dots,U_{i_n}\}\) is a finite subcollection that still covers all of \(\mathbb{R}\). Since every open cover has a finite subcover, \((\mathbb{R},\tau)\) is compact. Option (C) is TRUE.

Step 5: Check connectedness, option (D).
Suppose, for contradiction, \(\mathbb{R} = U \cup V\) with \(U,V\) disjoint, nonempty, and open. Since \(U \cap V = \emptyset\), \(V \subseteq U^c\), and \(U^c\) is finite (as \(U\) is open and nonempty), so \(V\) is finite. But \(V\) is also open and nonempty, so \(V^c\) is finite too. Then \(\mathbb{R} = V \cup V^c\) would be a union of two finite sets, making \(\mathbb{R}\) finite, which is false. So no such split exists, and \((\mathbb{R},\tau)\) is connected. Option (D) is TRUE.

Final Answer:
The space fails to be Hausdorff, but every finite set is closed, the space is compact, and the space is connected.
\[ \boxed{\text{(B), (C) and (D)}} \]