This probably will look like I'm trying to get you to answer my homework, but I'm not. All I'm looking for is to understand the problem and concepts involved with the problem. Here is the problems.

In the following problems in all metric subspaces $X$ of $\Bbb R$ the understood metric on $X$ is the standard metric of $\Bbb R$.

  1. Let $X = [0,1) ∪ (1,2]$. prove that the subsets $[0,1)$ and $(1,2]$ are both ”clopen” (meaning closed and open at the same time). Hint: use the theorem proved in class characterizing open and closed subsets of subspaces of metric spaces).

Let $E = [0,1)$

Since $E = X\cap(-1,1)$, E is open. Because (-1,1) is open. Thus proving (1,2] is closed because if the E is open, it's compliment is closed.

Now, let $G = (1,2]$

$G = X \cap (1,3)$ Thus proving that G is open, thus proving that E is closed. Both E and G are clopen. QED

  1. Consider the metric space $\Bbb Z$. Prove that any subset of $\Bbb Z$ is both open and closed in $\Bbb Z$.

Can someone guide me through this?

I know to be a closed set, a set must contain all of it's limit points. To be an open set, a set must contain all it;s interior points. But I can't find the theorem my professor talks about in my real analysis text

  • $\begingroup$ Could the theorem proven in class be that a subset of a metric subspace is open/closed when it is the intersection of an open/closed set in the underlying metric space and the subspace? $\endgroup$ – robjohn Oct 9 '14 at 20:05
  • $\begingroup$ That is it. Yall heloed to remind me $\endgroup$ – J.r Oct 9 '14 at 20:51

In a topological subspace $X$ of $\mathbb{R}$, a set $A$ is open if and only if it can be written as $A=X\cap O$, where $O$ is an open subset of $\mathbb{R}$.

Now think of your sets in the first problem. Can you write them in such a way? What about the complement of each of these?

For the second one, you can look at limit points. What does it mean if a sequence on $\mathbb{Z}$ converges? What does $U_\varepsilon(x)$ look like on $\mathbb{Z}$?


I suggest that you find an open set $U$ in $R$ whose intersection with $X$ is $[0, 1)$; that proves that $[0, 1)$ is open in $X$. (By the definition of open and closed sets in subspace topology.)

That should get you started for both problems. Forget the theorem. :)

And another hint: A closed set is one whose complement, in $X$, is open (in $X$). So once you've proved that $[0, 1)$ is open in $X$, you'll know that $(1, 2]$ is closed in $X$>

  • $\begingroup$ Ahhh okay. So would this be sufficient. $\endgroup$ – J.r Oct 9 '14 at 20:32
  • $\begingroup$ Let $E = [0,1) $ Since E = X intersect (-1,1), E is open. (And by consequence of that, this proves that the set (1,2] in X is closed because the compliment of an open set is a closed set). Now, to prove that [0,1) is closed, we do the same thing? $\endgroup$ – J.r Oct 9 '14 at 20:44
  • $\begingroup$ Dead right! You're on your way. $\endgroup$ – John Hughes Oct 9 '14 at 20:57
  • $\begingroup$ You are awesome. Thanks so much for the help. $\endgroup$ – J.r Oct 9 '14 at 21:00

For the first one, Use that $X$ is clopen and that the sets don't share any limit points.

For the second one, let $x\in Z$. Then consider any neighborhood, $N_r(x)$, of radius less than 1 around x. Then $N_r(x)\in \{x\}$ (in fact we have equality), hence any singleton sets is open. Moreover we know that any set in $Z$ is the (countable) union of singletons so it must also be open.

Do the same thing (almost) for the closed property.


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