Question

Let \((\mathbb{R}, \tau)\) be a topological space, where the topology \(\tau\) is defined as \[ \tau = \{U \subseteq \mathbb{R} : U = \emptyset \text{ or } 1 \in U\}. \] Which of the following statements is/are correct?

• A. \((\mathbb{R}, \tau)\) is first countable
• B. \((\mathbb{R}, \tau)\) is Hausdorff
• C. \((\mathbb{R}, \tau)\) is separable
• D. The closure of (1,5) is [1,5]

Hint:

For the included point topology on a set X with special point p (here, \(\mathbb{R}\) and 1), remember these rules of thumb:
- It is never Hausdorff (unless X has only one point).
- It is always separable, as \(\{p\}\) is a dense subset.
- The closure of a set A is \(A\) if \(p \notin A\), and it is X if \(p \in A\).

Answer 1

The Correct Option is A, C

Step 1: Understanding the Concept:
This problem asks us to analyze the properties of a specific topology on \(\mathbb{R}\), known as the "included point topology" with the point 1. The open sets are the empty set and any set containing the number 1. We need to check the standard topological properties of first countability, Hausdorff, separability, and compute a closure.
Step 3: Detailed Explanation:
(A) First countable: A space is first countable if every point has a countable local basis.
- For the point \(x=1\): The set \(\{1\}\) is open because \(1 \in \{1\}\). Any other open set V containing 1 must have \(\{1\} \subseteq V\). Therefore, the collection \(\mathcal{B}_1 = \{\{1\}\}\) is a local basis for the point 1. It is finite, hence countable.
- For any point \(x \neq 1\): Let V be any open set containing x. By definition of \(\tau\), we must have \(1 \in V\). Consider the set \(U_x = \{1, x\}\). This set is open because \(1 \in U_x\). Also, \(x \in U_x\) and \(U_x \subseteq V\). This holds for any open set V containing x. Therefore, the collection \(\mathcal{B}_x = \{\{1, x\}\}\) is a local basis for x. It is finite, hence countable.
Since every point has a countable local basis, the space is first countable. Thus, (A) is TRUE.
(B) Hausdorff (\(T_2\)): A space is Hausdorff if for any two distinct points \(x, y\), there exist disjoint open sets U containing x and V containing y.
- Let \(x=2\) and \(y=3\). Let U be any open set containing 2. By definition, \(1 \in U\). Let V be any open set containing 3. By definition, \(1 \in V\).
- This means \(1 \in U \cap V\), so the intersection is never empty. It is impossible to find disjoint open sets for any two distinct points.
- Therefore, the space is not Hausdorff. Thus, (B) is FALSE.
(C) Separable: A space is separable if it has a countable dense subset. A subset A is dense if its closure \(\bar{A}\) equals the entire space \(\mathbb{R}\).
- Let's determine the closed sets. A set C is closed if its complement \(\mathbb{R} \setminus C\) is open. The open sets are \(\emptyset\) and sets containing 1.
- So, \(\mathbb{R} \setminus C = \emptyset \implies C = \mathbb{R}\).
- Or, \(1 \in (\mathbb{R} \setminus C) \implies 1 \notin C\).
- Thus, the closed sets are \(\mathbb{R}\) and any subset of \(\mathbb{R}\) that does not contain the point 1.
- The closure of a set A, \(\bar{A}\), is the smallest closed set containing A.
- Consider the singleton set \(A = \{1\}\). Any closed set containing A cannot be a set that excludes 1. The only closed set containing A is \(\mathbb{R}\) itself.
- So, \(\overline{\{1\}} = \mathbb{R}\).
- The set \(\{1\}\) is a countable set whose closure is the entire space. Therefore, the space is separable. Thus, (C) is TRUE.
(D) The closure of (1,5) is [1,5]:
- Let \(A = (1,5)\). The point 1 is not in A.
- From our analysis above, any set that does not contain 1 is a closed set.
- Since \(1 \notin (1,5)\), the set \((1,5)\) is itself a closed set.
- The closure of a set is the smallest closed set containing it. Since \((1,5)\) is already closed, its closure is itself.
- \(\overline{(1,5)} = (1,5)\).
- The statement claims the closure is \([1,5]\). Thus, (D) is FALSE.
Step 4: Final Answer:
The correct statements are (A) and (C).