Bounded vs. unbounded, closed vs. open sets

Question

I am looking for some examples / definitions of these concepts so I can better understand other ones.

Namely, I am looking for all four permutations:

1. Unbounded and closed

2. Unbounded and open

3. Bound and closed.

4. Bound and open.

Apparently this is also related to the concept of "compactness" but "every open cover of the space has a finite subcover" doesn't mean anything to me.

Seeing some examples and understanding how these are defined would be helpful (I am trying to eventually understand the intuition behind the proof of Extreme Value Theorem but I need to understand these concepts first).

Answer 1

Sorry for two answers but in response to your comment:

"when I google these concepts it's too abstract, too vague,"

and

" I thought closed meant [a, b] but then we have [7, infinity) as closed."

I hope I can give an intuitive idea of "open" and "closed" and "bounded" and although in formal, tha is not misleading, and that in conjunction with a formal definition will render the formal definition less vague.

We have a universal space $X$. And for every point of $x$ we can draw little balls around this point. These balls can be as tiny as we want.

A set is a collection of points. Set can be connected or separate or any set of points.

A set $A$ is open: if for every point $a$ of $A$ we can draw a tiny ball around $a$ and all the points in the tiny ball will be in $A$.

So for example if $A = (0, 1)$ and we pick a point in $A$, say $a = 0.000001$. We can draw a tiny ball around $0.000001$ so that all points in the ball are in $A$. Because $a$ is so close to zero, the radius of this tiny ball must be very small. In fact it must be smaller than $0.0000005$. But we can do a ball and we can do a ball around every point. So $A$ is open.

As every point will have a ball around it, that means for every point there "a little further you can go". So $A$ will not have any sharp edges.

Now it is possible that there is a point $x$, (or more points) in the space $X$ that may, or may not be in $A$ and has the property, that every ball around $x$ will have some points in $A$.

For instance, take $A= (0,1)$ and $x = 0$. Notice every ball around $0$ will have some points that are bigger than $0$ (and smaller than $1$) and these will be in $A$. That will happen for every ball around $0$. It will also happen for every ball around $1$. And it will also happen for every ball around $0.75$ (or any point $y$ so that $0\le y \le 1$).

It will not happen for the point $-0.000001$ because we can take a tiny ball around $-0.0000001$ that completely misses $A$. But no ball around $0$ or $1$ can miss $A$.

We call such points "limit points" of $A$. Limit point: If every ball around $x$ must "hit" the set $A$ then $x$ is a limit point of $A$.

Notice $0.75$ is a limit point of $(0,1)$ and $0.75$ is in $(0,1)$ and notice $0,1$ are limit points of $(0,1)$ and $0$ and $1$ are not in $(0,1)$. If we were to list all the limit points of $A= (0,1)$ we would find they are the point $0$ and $1$ and all the points in between.

A set is closed: All its limit points are also points in the set.

So $(0,1)$ is not closed. Because the limit points $0$ and $1$ are not in the set.

But the set $B=[0,1]$ is closed. The only points that must hit $B$ when we draw balls around them are the points $[0,1]$, and those points are all in $B$.

Notice the space $X$ is both open and closed. For every point of $X$, every ball will be entirely in $X$ (there's nowhere else to be) so $X$ is open. Every ball around every point "hits" $X$ so every point is a limit point of $X$ and every point is in $X$ so $X$ is closed.

Notice not all sets have limit points. Take $\mathbb Z\subset \mathbb R$. If you take any real number you can draw a small ball around it and not hit any integer. So no point is a limit point of $\mathbb Z$.

This actually means that $\mathbb Z$ is closed. It doesn't have any limit points so all the limit points (all zero of them) are in the set so.... it is closed. It might be easier to say, there aren't any limit points that are not in $\mathbb Z$.

Notice the empty set is both closed and open. No points have balls that hit the empty set (there is nothing to hit) so there aren't any limit points of the empty set. So there aren't any limit points that are not in the empty set. So the empty set is closed.

The "empty set is closed" is a little more abstract. There aren't any points in the empty set that you can't draw a ball around and be entirely in the empty set. So for all the points that are in the empty st (all zero of them) you can't claim you *can't draw a ball around them them that is entirely in the empty set So the empty set is open.

Okay, bounded.... Well, bounded is exactly what it sounds like. Any distance between two points is finite. I'm not sure there is really anything more to say. The empty set is bounded because there is no distance between any two points.

Answer 2

This depends on context, but if you want intuition, working over the real line should be good. Here are some examples:

1. Unbounded and closed: $\mathbb Z$, $\mathbb R$, $[7,\infty)$.
2. Unbounded and open: $\mathbb R$, $\mathbb R\setminus\mathbb Z$, $(3,\infty)$.
3. Bounded and closed: any finite set, $[-2,4]$.
4. Bounded and open: $\emptyset$, $(0,1)$.

To check that these examples have the correct properties, go through the definitions of boundedness, openness, and closedness carefully for each set. Applying definitions to examples is a great way to build intuition. For example, some sets are both open and closed, but most are neither — sets are not doors.

On the real line compactness (every open cover has a finite subcover) is indeed equivalent with being bounded and closed. But that is by no means a tirival observation, and you should first build intuition to basic topological concepts like the ones you listed.

Answer 3

Here are some examples.

Unbounded and closed : The set of positive integers, $ \mathbb N $

Unbounded and open : The set of real numbers, $\mathbb R $

Bound and closed. : The closed interval , $[0,1]$

Bound and open.: The open interval, $(0,1)$

Compact: Closed and bounded sets or real numbers are compact: $[0,1]$, $\{1,2,34,5\}$

Answer 4

In $\mathbb R$ examples:

Unbounded and closed

All of $\mathbb R$. $(-\infty, 0]$, $[5,\infty)$, $[0,1]\cup [3,\infty)$. $[0,1]\cup \{2\}\cup [3,\infty)$, $\mathbb Z$.

Unbounded and open

All of $\mathbb R$, $(-\infty, 0)$, $(5,\infty)$, $(0,1)\cup (3,\infty)$.

Bound and closed.

$\emptyset$, $[-2,0]$, $[5,7]$,$ [-2,0] \cup [0,1] \cup [3,5]$.

Bound and open.

$\emptyset$, $(-2,0)$, $(5,7)$, $(-2,0) \cup (0,1) \cup (3,5)$.

Apparently this is also related to the concept of "compactness" but "every open cover of the space has a finite subcover" doesn't mean anything to me.

Oh boy, do I feel for you.

Here is an intuitive example. It's very soft and informal.

Take $A = [0,1]$ and $B = (0,1)$. $A$ is closed and bounded. It's also compact. $B$ is open and bounded but isn't compact.

Let's try to cover these up with open intervals.

To start with $A$ we need to cover up the spot $0$ with and open interval $(a,b)$ where $a <0$ and $b > 0$. And we can choose this interval to be really small. As small as we like. (In fact we want it to be small because we want to thwart ourselves.)

So that interval covers $[0,b)$ and doesn't cover $[b,1]$. We now need a second interval $(c,d)$ where $c < b < d$. Again we can make this as small as we want but it must "go past" $b$.

No we can do this forever and never actually get to the end and cover the spot $1$. But if we have a rule that we must eventually take the "final bite" and cover the spot $1$. We will have done it in a finite number of intervals.

Summary: To cover all of $[0,1]$ there have to have been a first and final bite and a finite number of bites in between. In other words for any cover of open intervals that cover all of $[0,1]$ we can pick out a finite number of the inervals to cover.

Now $B$ is different. We don't have to cover $0$ or $1$. Just all the points up to them. So we don't need to take any "first bites" or "last bites". We can take an infinite number of nibbles up to the edges.

This is like the flea hopping across the desk. Each leap he jumps half way to the edge. In this way he can get as close to the edge but never get to the edge. In an infinite number of leaps he'll cover all the points except the edge.

But if we only take a finite number of his leaps we can only get to $\frac{2^n-1}{2^n}$ and all the point beyond are not reached.

But if we say "one of his leaps must get to the edge" and that leap is $1-teenyleap$ to $1$. Then there were a finite number $n$ of leaps where he got to $\frac {2^-1}{2^n}$ and $1-teenyleap < \frac {2^-1}{2^n}$.

If one of his leaps must go to the edge, then no matter how we arrange them we can pick a finite number of leaps to do it in. But if the edge need never be reached but can be a distant unattainable goal, we can cover the desk in a required infinite number of leaps.

The thing is by have us need to take all the points between $0$ and $1$ but not $0$ and $1$ we can do it in an infinite number of nibbles where all nibbles are required. If we only take a finite number of nibbles we can leave out a huge number of spots.

But by forcing us to say, one of our nibbles must include $0$ and another must include $1$, those two bites will make all but a finite number of our nibbles unnecessary.