Determine if a Set is Open or Closed

[hammonjj]

Homework Statement

Determine each of the following sets as open, closed, neither or both.
a) {1/n : n $\in$ $N$ }

b) $N$

c) $Q$

d) $\bigcap$$^{∞}$$_{n=1}$(0,1/n)

e) {x: |x-5|$\leq$ 1/2

f) {x: x^2>0}

Homework Equations

Open sets are sets that do not contain their boundary points. Closed sets contain their boundary points. Also, the compliment of an open set is closed and the compliment of a closed set is open.

The Attempt at a Solution

a) Closed because the natural numbers are closed.

b) Naturals are closed because each neighborhood contains only the Natural number (ie. the natural number is both the interior and boundary point)

c) Q is neither open nor closed.

d) (0,1/n) is closed for the same reasons as part a and the intersection of any number of closed sets is closed.

e) Closed because +/- of 1/2 is contained within the interval.

f) Not sure, 0 is not in the interval because x^2 is strictly greater than 0, but the set continues on to infinity. Does that make the interval open because you can't make a neighborhood around infinity?

Thanks a bunch for the help on this! We're just getting into topology of the Reals and I want to make sure I've got this right before I move on to some of the more complicated proofs.

 

[lanedance]

with these questions you need to be careful with terminology and some definitions... In particular you need to be clear which "universe" you're working in, so I assume you're asking which of these sets is closed in the set R? (correct me if I'm wrong..)

a) the fact the natural numbers are closed in does not imply this set is closed. Consider an open neighbourhood of the origin...
b) true for any neighbourhood with r<1
c) do you need to explain this?
d) (0,1/n) is not a closed set, so I don't think your argument quite holds up. Try and have a think about the limiting behavior... what points are left in the intersection?
e) +-1/2 isn't in that interval, but 4.5 and 5.5 are
f) i think so, but its been a while