It is a closed set because it's the reciprocal image of a closed set by a cont. application.

Proof :

Let's show that the reciprocal image by a continuous application of an open set is open.

Let , where O is an open set. Because it's open, it's a neighbourhood of any of its point. Hence it is a neighbourhood of f(x).

Because the function is continuous, will be a neighbourhood of x.

Hence , is a neighbourhood of x. So it's a neighbourhood of all its points. So it's an open set.

Now use Hausdorff's formula :

is a closed set.

is open, and therefore is a closed set.

We're done.

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and in the usual topology over R, {0} is a closed set.

No, but they could ask you to show it is a compact.Aside: (In the past we have used K to denote a compact set. I'm not sure if it is denoting one here or not. Would they have to specify in the beginning of the problem that K is compact?)