1. ## New Question

A subset A of a metric space X is called uniformly discrete if there
epsilon> 0 with the property that the distance between two distinct points of A is always
greater or equal to epsilon:
∀ a, b ∈ A, (a!= b) ⇒ (dX (a, b) ≥ epsilon).
Show that any subset of a uniformly discrete metric space X is
closed in X.

2. Originally Posted by donsmith
A subset A of a metric space X is called uniformly discrete if there
epsilon> 0 with the property that the distance between two distinct points of A is always
greater or equal to epsilon:
∀ a, b ∈ A, (a!= b) ⇒ (dX (a, b) ≥ epsilon).
Show that any subset of a uniformly discrete metric space X is
closed in X.
Always begin a thread for a new question.

Say tha $\displaystyle \varepsilon$ is fixed by the problem.
If $\displaystyle x$ were a limit of some set $\displaystyle Y$ then the ball $\displaystyle B\left( {x;\frac{\varepsilon }{2}} \right)$ must contain a point of $\displaystyle Y$ distinct from $\displaystyle x$.
But that is impossible.
So $\displaystyle Y$ is closed by default.

3. solved