# Topology of the Reals

1. Sep 6, 2011

### Shackleford

Can I simply combine the unions into a single interval like I did in (a)? The closed interval [1,4] fills in the (3) hole from (0,3), etc. I did something similar in (b).

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110906_221620.jpg [Broken]

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110906_221628.jpg [Broken]

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110906_221635.jpg [Broken]

Last edited by a moderator: May 5, 2017
2. Sep 6, 2011

### Shackleford

Yeah, I just want to make sure my answers are correct.

I have the correct interior, boundary, accumulation, and isolated points for (a), (b), and (c)? I didn't have any isolated points.

3. Sep 6, 2011

### lanedance

sorry its not exactly clear what you're attempting?

also it would probably be quicker to type these, that way I could cut and paste as well - and I'm lazy

you need to find the
- int = interior?
- bd = boundary?
- Sbar = closure?

of teh given set based on the "normal" topology of the reals?

4. Sep 6, 2011

### lanedance

ok whats accumulation? is that accumulation points, so all limit points or the closure of S?

5. Sep 6, 2011

### lanedance

ok, so starting again a) looks good

6. Sep 6, 2011

b) also

7. Sep 6, 2011

### SammyS

Staff Emeritus
They look right.

Took me a while to decide on (c). The thread title helped.

I assume it's the Euclidean (usual) Topology on the reals.

8. Sep 6, 2011

### lanedance

c) not so much

for this case, the interior is not the open interval as S does not contain irrational numbers...

i think the closure and boundary are ok as S is dense in the reals and also in (0, sqrt(2))

now what is your definition of isolated?

9. Sep 7, 2011

### SammyS

Staff Emeritus
I am in error regarding (c).

No neighborhood of any point in S is contained in S.

10. Sep 7, 2011

### Shackleford

Yeah, for (c), I wasn't sure because you have rationals and irrationals mixed in the set. Could I set the interior points to be open interval minus the irrational numbers? Maybe I could do the open interval intersect with the rationals.

An isolated point is a point that's in S but not an accumulation point.

11. Sep 7, 2011

### lanedance

the way i read it, there's only rationals in S

for the interior how about just S? there's nothing to say it can't equal its interior (unless you can find an issue)

i think the rest is ok

12. Sep 7, 2011

### SammyS

Staff Emeritus
For (c):

Any open set that contains a point of S, also contains points not in S, so it is not a subset of S.

It seems to me that the interior of S is empty.

13. Sep 7, 2011

### micromass

Staff Emeritus
for (c): the interior of S is empty, the boundary of S is $[0,\sqrt{2}]$. Can you figure out why?

14. Sep 7, 2011

### lanedance

yep agree

15. Sep 9, 2011

### Shackleford

Well, for each rational point in the interval, you can find irrational numbers in every neighborhood, and so the intersection with the rationals and irrationals (S and S-complement) is always nonempty.

Every point is either an interior or boundary point, and since every point in S is a boundary point, that means int S equals the empty set.

By the same reasoning, each boundary point $[0,\sqrt{2}]$ is also an accumulation point. You can always find rationals in all of your deleted neighborhoods. Since there are no interior points, each point is an isolated point. But that is contradictory.

Last edited: Sep 10, 2011
16. Sep 11, 2011

### Shackleford

I'm reasonably confident these answers are correct.

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110911_113312.jpg [Broken]
http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110911_113318.jpg [Broken]
http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110911_113323.jpg [Broken]

Last edited by a moderator: May 5, 2017