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## Main Question or Discussion Point

So I've got this homework problem that I think I've found a counter-example to, so either my counter-example is wrong or the professor is, here is the problem:

Let (X,d) be a metric space and let E be a nonempty subset of X. For each x ∈ X, let d(x,E) = inf{d(x,y) : y ∈ E, with y≠x}.

Show that {x ∈ X : d(x,E)<r} is open for each r ∈ ℝ.

So now for my counter-example:

Let (X,d) = {1,2,4} with the usual metric and ordering and define our topology τ on (X,d) to be τ = {∅, {1}, {4}, {1,2,4}, {1,4}}. Now simply take E = X and notice that d(1,E)=1, d(2,E)=1, and d(4,E)=2. Thus take r=1.5 and note that this means that {x ∈ X : d(x,E)<1.5} = {1,2}, which is closed according to our topology τ.

Help would be much appreciated, thanks =].

Let (X,d) be a metric space and let E be a nonempty subset of X. For each x ∈ X, let d(x,E) = inf{d(x,y) : y ∈ E, with y≠x}.

Show that {x ∈ X : d(x,E)<r} is open for each r ∈ ℝ.

So now for my counter-example:

Let (X,d) = {1,2,4} with the usual metric and ordering and define our topology τ on (X,d) to be τ = {∅, {1}, {4}, {1,2,4}, {1,4}}. Now simply take E = X and notice that d(1,E)=1, d(2,E)=1, and d(4,E)=2. Thus take r=1.5 and note that this means that {x ∈ X : d(x,E)<1.5} = {1,2}, which is closed according to our topology τ.

Help would be much appreciated, thanks =].