general-topologyreal-analysisvector analysis
$$U = \{(x_1,x_2) \in \mathbb{R}^2 : -1 < x_2 \leq 1\}$$
(a) Find all interior points of U. For each interior point, find a value of r for which the open ball lies inside U.
(b) Find all boundary points of U.
(c) Is U an open set? Is U a closed set? Why or why not?
So I know the definitions of boundary points and interior points but I'm not sure how to use the definitions of each to solve for this. First time attempting a problem like this, any guidance would be appreciated.
A boundary point of a set $A$ cannot be an interior point, but that’s not a problem: the theorem is that the closure of $A$ consists of those points that are in the interior of $A$ together with those that are in the boundary of $A$. In symbols,
$$\operatorname{cl}A=\operatorname{int}A\cup\operatorname{bdry}A\;,\tag{1}$$
and in fact the two sets on the righthand side of $(1)$ are disjoint. (By the way, a closed set need not have any boundary points at all: in $\Bbb R$ the only examples of this phenomenon are the closed sets $\varnothing$ and $\Bbb R$, but in more general topological spaces there can be many sets that are simultaneously open and closed and which therefore have empty boundary.)
You know that $x\in\operatorname{int}A$ if and only if $x$ has an open nbhd contained in $A$, and that $x\in\operatorname{bdry}A$ if and only if every open nbhd of $x$ intersects both $A$ and $\Bbb R\setminus A$. To prove $(1)$, I suggest that you first prove that
$\qquad\qquad\qquad\quad x\in\operatorname{cl}A$ if and only if for each open nbhd $V$ of $x$, $V\cap A\ne\varnothing$.
Once you have this, $(1)$ is pretty straightforward.
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
A drawing of the set U with interior and boundary points
The picture is a drawing of the set U (drawn in purple). The dashed line at $x_2=-1$ is not included in the set U, while the solid line $x_2=1$ is contained in U. Moreover, any point with $x_2$ strictly between -1 and 1 is contained in U. There are three points highlighted inside of U, namely w, y, and z.
Intuitively, y is in the interior of U. In order to show this mathematically, you can use the definition of an interior point that some open ball around y is contained in U. A ball of radius $\epsilon>0$ around y is of the form $B(y; \epsilon)=\{x\in \mathbb{R}^2: \|x-y\|<\epsilon\}$ (in the picture, I drew such a ball). So if $y\in U$ is chosen so that $y_2\in (-1, 1)$, then you can choose $\epsilon$ small enough so that $B(y; \epsilon)\subseteq U$. You'll have to do a bit of work to find such an $\epsilon$.
On the other hand, w and z are on the boundary of U. To show this mathematically, you can use the definition of a boundary point showing that every ball around w or z touches a point in U and a point outside of U. Consider some ball $B(w; \epsilon)$ around $w$ with radius $\epsilon>0$ (in the picture, I drew an example). Note that, since $w_2 = -1$, any choice of $\epsilon$ will result in some point $q\in B(w; \epsilon)$ with $q_2 <-1$ and some point $r\in B(w; \epsilon)$ with $r_2>-1$. Hence $q\not\in U$ but $r\in U$. This shows that $w$ is on the boundary.