Which sets are open and closed in a subspace?

[Damascus Road]

Here's two more question I'm working on in test prep.

2.) Let Y = [-1,1] have the standard topology. Which of the following sets are open in Y, and which are open in R.

A= (1,1/2) $$\cup$$ (-1/2,-1)
B= (1,1/2] $$\cup$$ [-1/2,-1)
C= [1,1/2) $$\cup$$ (-1/2,-1]
D= [1,1/2] $$\cup$$ [-1/2,-1]
E= $$\cup$$ $$\frac{1}{1+n}, \frac{1}{n}$$ (union is from n=1 to infinity)

So, to start, I should I examine their complements?
My gut feeling is that A,B are open in R and C, D are open in Y and I'm not sure about E. 3.) I need to prove:
Let X be a topological space, and let Y $$\subset$$ X have the subspace topology. Then C $$\subset$$ Y is closed in Y iff C = D$$\cap$$ Y for some closed set D in X.

This has be a bit confused...
A subspace topology on Y is defined as
[tex] T_{Y} = {U \bigcup Y | U is open in X} [\tex]

So, simply, if D were open... it's basically the exact definition that I provided. Which gives that U is open.

 

[lanedance]

exmining their complements could be uesful in some cases but in others it should be resonably clear

a good example is 2)d)
[-1,1] is open in Y as it is Y, but clearly closed in R

 

[lanedance]

for 3) your definition of the subspace topology is not quite correct it should be an intersection
$$T_{Y} = {U \cap Y | U \ is \ open \ in \ X}$$

 

[SammyS]

In 2., is set E a union of open intervals, or a union of closed intervals?

 

[lanedance]

reasonable for the others but in d) is is a union of half open intervals in Y, but as they cover Y, they results in an open set as Y is open in itself