let X be a metric space. show that every subset of X is open if and only if each subset of X which consists of a single point is open

Let's prove that in a metric space \( X \), every subset of \( X \) is open if and only if each subset of \( X \) which consists of a single point is open.

### Part 1: If Every Subset of \( X \) is Open, Then Each Single-Point Subset is Open

**Given:** Every subset of \( X \) is open.

**To Show:** Each single-point subset \( \{x\} \) for \( x \in X \) is open.

**Proof:**
- By assumption, every subset of \( X \) is open. 
- A single-point subset \( \{x\} \) for any \( x \in X \) is a subset of \( X \).
- Therefore, \( \{x\} \) must be open because all subsets are open by assumption.

Thus, if every subset of \( X \) is open, then each subset consisting of a single point is open.

### Part 2: If Each Single-Point Subset of \( X \) is Open, Then Every Subset of \( X \) is Open

**Given:** Each single-point subset \( \{x\} \) for \( x \in X \) is open.

**To Show:** Every subset of \( X \) is open.

**Proof:**
- Let \( A \subseteq X \) be any subset of \( X \).
- We want to show that \( A \) is open.
- Consider any point \( a \in A \). We need to find a neighborhood around \( a \) that is entirely contained within \( A \).
- Since each single-point subset \( \{a\} \) is open, by definition, there exists some \( \epsilon > 0 \) such that the open ball \( B(a, \epsilon) \subseteq \{a\} \).
- However, the open ball \( B(a, \epsilon) \) containing only \( a \) implies \( B(a, \epsilon) = \{a\} \), which is a contradiction unless the space is discrete.
- Thus, for this to hold in a non-trivial metric space, \( X \) must be discrete, meaning that every point is isolated.
- In a discrete metric space, every subset is open because for any point \( x \in X \), the single-point subset \( \{x\} \) is open, and any subset of a discrete space is a union of open sets (hence open itself).

Thus, if each single-point subset of \( X \) is open, then every subset of \( X \) is open.

### Conclusion
We've shown that every subset of \( X \) is open if and only if each subset of \( X \) consisting of a single point is open. This result essentially characterizes discrete metric spaces, where the only case this equivalence holds is when \( X \) is discrete.

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Would you like more details on any part of the proof or have any other questions? Here are some related questions to explore further:

1. What is the definition of an open set in a metric space?
2. How do discrete metric spaces differ from other metric spaces?
3. Can you give an example of a non-discrete metric space where not every subset is open?
4. What are some properties of metric spaces where every point is an isolated point?
5. How does this proof extend to topological spaces, beyond metric spaces?

**Tip:** Understanding the concept of discrete spaces helps in identifying when each singleton set being open implies every set is open.

Explore the proof that in metric spaces, every subset is open if and only if each subset consisting of a single point is open. Understand the implications for discrete metric spaces and the definition of open sets. Suitable for advanced undergraduate level students and those interested in topology.

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