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  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by donsmith View Post
    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.
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