Exercise 1.10

Show the following: In the standard metric on the line $$[0,1)$$ is neither open nor closed.

Unconfirmed Solutions
For any ball $$B(0, \varepsilon) = \{ x| x + \varepsilon < 0 < x - \varepsilon \}$$ there is some $$a \in B(0, \varepsilon), \text{ and } x < -\varepsilon$$ and thus $$B(0, \varepsilon) \not\subset [0,1)$$, hence [0, 1) is not open. Consider its compliment $$(-\infty, 0) \cup [1, \infty) = A$$ and repeat the above proof for any ball $$B(1, \varepsilon)$$ to see that $$A$$ is not open and therefore its compliment is not closed.

baruch_shahi
Consider the ball $$B(0,r)$$ centered at the origin with radius $$r$$. Certainly this ball contains points of $$\mathbb{R}$$ not contained in $$[0,1)$$; for example, $$-\frac{r}{2}$$. Thus $$B(0,r)$$ is not completely contained in $$[0,1)$$ for any choice of radius $$r$$. Therefore, $$[0,1)$$ is not open in $$\mathbb{R}$$ with the usual metric.

Now consider the complement $$\mathbb{R}\setminus [0,1)=(-\infty, 0)\cup [1,\infty)$$. Using the work above, it should be clear that this set also is not open. For example, consider the ball $$B(1,r)$$ for any radius $$r$$. Since its complement is not open, $$[0,1)$$ is not closed.

baruch_shahi 21:21, February 27, 2011 (UTC -5)

'''Absolutely correct. --Steven.clontz 16:06, March 1, 2011 (UTC)'''