Joy, I think it would be good to reflect on the difference between the quantifiers 'for all' and 'there exists', since this confusion seems to creep in in many places below. I'll now write some comments to your questions, hopefully they'll help clarify things.
<1>My understanding for interior point is that:
if we have a open set A in ℝd, then every ball (centered at x
with radius r that we draw are all completely contained in this open set. If it is in ℝ, then we think this in terms of real line instead of balls.
It doesn't have to be true for every ball. For example, $0$ is an interior point of $(-1,1)$, but the open ball $B_3(0)$ of radius $3$ around $0$ ($B_3(0) = (-3,3)$) isn't contained in $(-1,1)$. The correct thing to require is that there exists an open ball of some radius around $x$ that is entirely contained in $A$. So for the example with $0$ in $(-1,1)$, we see that $0 \in (-1/2, 1/2) = B_{1/2}(0) \subset (-1,1)$, which is enough to conclude that $0$ is an interior point (many other choices of interval/radius would work just as well).
Also, the case of $\mathbb{R}^1$ isn't really different from $\mathbb{R}^d$, it's just that `balls' in one dimension look like intervals.
For this question, it has no interior points because if we write out this sequence, we can actually see that this is a rational sequence, and if we take an a small interval over this sequence, it will contain both rational number and irrational number, so it does not have interior point.
This is a nice argument, but we should be considering each point individually, not taking a `small interval over this sequence'. But this kind of argument will work for every point of $A$ (any nonempty ball around a rational point will contain irrational points, so can't be contained entirely in $A$).
<2> For cluster point:
If we draw a real line and have our interval, let's say $[a,b]$, we pick a point $A$ from the interval and draw two other small intervals, let's say $[c,d]$, $[e,f]$. We check if these two small intervals $[c,d]$, $[e,f]$
contain the points in our big interval $[a,b]$.
For this question, I think the cluster point is the real number, because the numbers we get will include both rational and irrational number, and they are contained in the real numbers ℝ.
I don't follow what you mean. According to the definition on p.51 of
http://www.mast.queensu.ca/~andrew/teaching/math281/pdf/281-reader.pdf
a point $x$ is called a cluster point of $A$ if the punctured ball around $x$ of any radius $r>0$ contains a point of $A$. Another way to say this is that for any $r>0$, the open ball $B_r(x)$ contains a point of $A$ other than $x$ (to get the punctured ball, we delete $x$ from $B_r(x)$).
It's definitely not the case that every point of $\mathbb{R}$ is a cluster point of $A$. For example, the open ball $B_{1/3}(1)$ of radius $1/3$ about $1$ doesn't contain any points of $A$ except $1$; in other words the punctured ball of radius $1/3$ about $1$ is empty. So $1$ isn't a cluster point, and so on.
As a hint, I'll let you know that $A$ has exactly one cluster point. What would it be, and why is there only one?
<3> For Limit point. I am not quite sure about limit point, I always mix this up with the limit of a sequence. I am guess the limit point for this question is also real number, but I am just guessing.
The definition of limit point is similar to that of a cluster point. According to p.53 of
http://www.mast.queensu.ca/~andrew/teaching/math281/pdf/281-reader.pdf
a point $x$ is called a limit point of $A$ if any (non-empty) open ball around $x$ contains a point of $A$.
The difference is that we're no longer working with the punctured ball, so this point could be $x$ itself, as long as $x$ is a point of $A$.
What is the relationship with limits of sequences in $A$? If $x$ is a limit point of $A$, we can construct a sequence in $A$ that converges to $x$; conversely, if $(a_n)$ is a convergent sequence in $A$, then its limit is a limit point of $A$ (whether or not the limit is in $A$).
Every cluster point is a limit point, but the converse is not true. For example, if $A = [0,1] \cup \{2\}$, then $2$ is a limit point but not a cluster point of $A$.
Once again, the set of limit points is not all of $\mathbb{R}$ (what would be a sequence in $A$ that converged to $3 \in \mathbb{R}$, for instance?).
<4> isolated point: Points in A (a subset) and does not contain other points. If we have a real line, isolated points are these points that after we draw our small interval, there will be only one point (itself) in the interval, it does not contain any other points. (an example of isolated point maybe integers)
For this question, I think we do not have isolated points, since we can have infinite many terms in each small interval.
Right definition, but it's not true that there are no isolated points. In fact, there are many of them. Remember, that it's enough to find one open interval/ball of some radius that contains $x$ and doesn't contain any other points of $A$ to show that $x$ is isolated. It doesn't have to hold for every possible interval.
<5> boundary point: Is the boundary point as same as the endpoints? If we think in $\mathbb{R}^d$, boundary points are these points that are both contained in the subset A and its complement.
For this question, would the boundary point be $(1,0)$? I have this answer because after I write the sequence out, the first term is $1$ and this sequence will approach to $0$.
The boundary points of an interval like $[a,b]$, $(a,b]$ and so on will be its endpoints, but in general this isn't true.
Again, going to our notes, on p.54 we find that a point $x$ is a boundary point of $A$ if a (non-empty) open ball of any radius around $x$ contains points of both $A$ and the complement of $A$.
You already made the observation that $A$ doesn't contain any irrational numbers, while any nonempty open interval in $\mathbb{R}$ contains irrational numbers. This is enough to conclude that: ???
I think this is neither closed nor open set
Yes, but why?
Best Answer
I think you misread the meaning of $(\frac1{n+1},\frac1n)$: It is the open interval with endpoints $\frac1{n+1}$ and $\frac1n$.
The set is the union of open sets and hence open, which implies that all points of $X$ are interior. Indeed, if $x\in X$, then $x\in(\frac1{n+1},\frac1n)$ for some $n$, i.e. $\frac1{n+1}<x<\frac1n$. With $\epsilon=\min\{x-\frac1{n+1},\frac1n-x\}>0$ the $\epsilon$ ball around $x$ is in $X$.
The set of cluster point and the closure are both equal to $[0,1]$ as precisely the $x$ with $0\le x\le 1$ have at least one/infintely many other points in every open neighbourhood.
The boundary $\partial X$ is the difference of closure and interior, i.e. $\{\frac1n\mid n\in\mathbb N\}\cup\{0\}$.