Is $A$ open, closed, clopen, neither? Dense or not dense? The following are sample midterm problems with solutions:


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*$X=\mathbb{R^2},A=\{(x,nx):x\in\mathbb{R},n\in\mathbb{Z}\}$
Solution: Neither, not dense. 
Question: I don't understand why this set is not closed. I thought this would be all the lines passing through the origin with integer slope... so why is it not closed?

*$X=\mathbb{Z}^\mathbb{N},A=\{x\in X:x_n\in(1,2,\ldots,n)\forall n\in\mathbb{N}\}$
Solution Closed, not dense.
Question: This may sound silly, but what exactly is the topology on $\mathbb{Z}^\mathbb{N}$?

*$X=\mathbb{R}^2,A=\{(p,pq):p,q\in\mathbb{Q}\}$
Solution: Neither, dense.
I think I understand this because $A$ is just $\mathbb{Q}^2$

*$X=\mathbb{R}^2,A=\{(x,rx):x\in\mathbb{R}_{>0}, -1<r<1\}$
Solution: Open, not dense.
Since $A$ is the open cone between graph of $f(x)=x$ and $f(x)=-x$ in the right half plane.

*$X=\{0,1\}^\mathbb{N}=\{x=(x_0,x_1,\ldots):x_1,x_2,\ldots\in\{0,1\}\},A=\{x\in X:x_5=0\}$
Solution: clopen, not dense
Question: What is the topology on $\{0,1\}^\mathbb{N}$? Is it just discrete topology on $\{0,1\}$ in each coordinate so since any $x$ with $x_5=1$ has a neighborhood in that coordinate it is open...? Would this mean every sequence is a clopen set? 

*$X=\{0,1\}^\mathbb{N}, A$ is the set of $x\in X$ for which $x_n=1$ for at most finitely many $n\in\mathbb{N}$
Solution: Neither, not dense
I think getting the topology straight in question 5 will make this one trivial.
 A: I am going line by line through the problem. Apologies in advance for formatting errors.


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*$X=\mathbb{R^2},A=\{(x,nx):x\in\mathbb{R},n\in\mathbb{Z}\}$
Solution: Neither, not dense. 
Question: I don't understand why this set is not closed. I thought this would be all the lines passing through the origin with integer slope... so why is it not closed?
The vertical line through the origin (the $y$-axis with "slope infinity") is not included in this set, but every point on it is a limit point of points in this set.
The set is clearly not open or dense (it doesn't contain any balls and its closure misses a huge sector of the first quadrant for instance).


*$X=\mathbb{Z}^\mathbb{N},A=\{x\in X:x_n\in(1,2,\ldots,n)\forall n\in\mathbb{N}\}$
Solution Closed, not dense.
Question: This may sound silly, but what exactly is the topology on $\mathbb{Z}^\mathbb{N}$?
If it's not specified, then $\mathbb{Z}$ should get the discrete topology and any product should get the product topology. The product topology is the coarsest topology such that all the projection maps are continuous. In the product topology, a sequence $y_n \to y$ converges if and only if $\pi_i(y_n) \to y$ converges for each projection map $\pi$. From this it follows that your set is closed, and is not dense since it doesn't contain $(1, 2, 3, 4, 5, \ldots)$.


*$X=\mathbb{R}^2,A=\{(p,pq):p,q\in\mathbb{Q}\}$
Solution: Neither, dense.
Question: I think I understand this because $A$ is just $\mathbb{Q}^2$
You have the right idea, but aren't quite right: for example, $A$ doesn't contain $(0, 3)$.


*$X=\mathbb{R}^2,A=\{(x,rx):x\in\mathbb{R}_{>0}, -1<r<1\}$
Solution: Open, not dense.
Question: Since $A$ is the open cone between graph of $f(x)=x$ and $f(x)=-x$ in the right half plane.
I agree.


*$X=\{0,1\}^\mathbb{N}=\{x=(x_0,x_1,\ldots):x_1,x_2,\ldots\in\{0,1\}\},A=\{x\in X:x_5=0\}$
Solution: clopen, not dense
Question: What is the topology on $\{0,1\}^\mathbb{N}$? Is it just discrete topology on $\{0,1\}$ in each coordinate so since any $x$ with $x_5=1$ has a neighborhood in that coordinate it is open...? Would this mean every sequence is a clopen set? 
Again, you are taking the product topology, so, each coordinate gets the discrete topology and a sequence converges if and only if each of its projection converge. The easiest way to see this set is clopen is to view it as the inverse image of the clopen set $\{1\}$ under the projection map $\pi_5$.
The product topology is not discrete, even though all the factors are. For example, the sequence $(1, 0, 0, 0, \ldots), (0, 1, 0, 0, \ldots), (0, 0, 1, 0, \ldots),\ldots$ converges to $(0, 0, 0, \ldots)$.


*$X=\{0,1\}^\mathbb{N}, A$ is the set of $x\in X$ for which $x_n=1$ for at most finitely many $n\in\mathbb{N}$
Solution: Neither, not dense
Question: I think getting the topology straight in question 5 will make this one trivial.
It isn't closed; you can easily get $(1, 1, 1, \ldots)$ a a limit of elements of this set, e.g. by taking $x_n = (1, 1, 1, \ldots, 1, 1, 0, 0, \ldots)$ where the first $n$ elements are $1$. 
It isn't open. The complement is the set of sequences with infinitely many $1$s, which isn't closed, becuase you can get $(0, 0, 0, \ldots)$ as a limit by doing the complement of the above described procedure.
This set seems dense to me. What tuple can't I get as a limit by modifying the above procedure?
A: 3/ A={(p,pq)/ as pand q are in Q} 
We have A is included in QQ so interior(A) is included in interior(QQ)
And interior(Q*Q)=interior(Q)*interior(Q)=emptyset
So interior(A)=emptyset wich is different than A 
Conclusion: A is not an open 
To prove that A is not closed just think in term of sequences of A but their limite isnt in A 
