Exercise 1.10.b

Show that every subset of a discrete metric space is clopen (both open and closed).

Avidya
Let $$X$$ be a metric space with the discrete metric. For any $$C \subseteq X$$, every element $$ x \in C$$ has a neighborhood $$B(x,\frac{1}{2}) = \{x\} \subseteq C$$. Therefore, every $$C$$ is open.

Consequently, for any element $$y \in X \setminus C$$, $$B(y,\frac{1}{2}) = \{y\} \subseteq X \setminus C$$, therefore, every $$X \setminus C$$ is also open. By definition, this means $$C$$ is closed. Since any $$C$$ is both open and closed, any $$C$$ is clopen.

Still yet to rigorously prove: Clopen sounds ridiculous.

Avidya 17:57, March 3, 2011 (UTC)

'''Looks good. You could have mentioned in lieu of the second paragraph that since every set is open, the complement of every set is open, and therefore every set is closed. Also, clopen is an AWESOME word, but I guess a proof of it is independent of ZFC. :) --Steven.clontz 19:32, March 3, 2011 (UTC)'''

Spicyice
Let (X,d) be a metric space with d the discrete metric. Pick any subset S of X. Pick any epsilon e less than 1 and consider the open ball B(x_i,e) which then contains only the point x_i, and x_i is an element of S. But now B(x_i,e) is a subset of S, so we have found an epsilon neighbourhood around it. Therefore every point in S is an interior point. Therefore S is an open neighbourhood.

To show that S is closed consider any convergent sequence {x_n} n > 0. Since this sequence converges to some point t (not necessarily in S but we will prove that it is), for all epsilon e greater than 0 we can find an integer N such that for all n greater than N we have d(t, x_n) < e. Choose e to be any number less than 1. By definition of the discrete metric this means x_n = t. This means after a sufficient number of terms in the sequence, the sequence is comprised of just one number, and further this number is in S. Therefore since any convergent subsequence in S converges to a point in S we see that S holds all its limit points and therefore is closed.

'''The first part is true, but the second part uses a theorem that we haven't covered. (Sequences will be covered next week, but this theorem isn't listed there.) There is a much simpler solution for showing every set is closed (hint: look at the definition of closed from our notes). --Steven.clontz 22:19, March 3, 2011 (UTC)'''