Prove Set U is Open | Product Topology on R^2

[sa1988]

Homework Statement

Let $U = \{(x,y)\in R^2 : xy > 0\}$

Show that $U$ is open in the product topology on $R^2$ induced from the standard topology on $R$

Homework Equations

The Attempt at a Solution

Proofs are my downfall.

First I've visualised $U$ and it seems to be that it's essentially defined by the upper-right and lower-left quadrants of the x-y plane, not including the points on the axes themselves.

From what I understand, for the proof I need to show that every point $t \in U$ has an open neighbourhood, i.e the neighbourhood is entirely within $U$.

I believe the best way would be to assume $U$ is closed, then look for a contradiction.

So, assuming $U$ to be closed,

its compliment $U^c$ must be open.

But $U^c = \{ (x'y')\in R^2 : x'y' \leq 0 \}$

which has a clear limit at the point $x = 0$ or $y = 0$

so there are neighbourhoods $N$ around the points $n = (0,y')$ and $n = (x',0)$ where $n \in N$ but $n \notin U^c$

Hence $U^c$ must be closed, so $U$ is open.

 

[vela]

If $U^c$ is not open, it doesn't necessarily mean it's closed. In R, for example, the interval (0,1] is neither open nor closed.

I think a direct proof would be pretty straightforward. If $(x_0, y_0) \in U$, find an open set $N=(x_0-\epsilon,x_0+\epsilon)\times(y_0-\epsilon,y_0+\epsilon)$ where $\epsilon>0$ is appropriately chosen. It should be straightforward to show that every point in that set lies entirely in $U$.

 

[fresh_42]

I believe the best way would be to assume $U$ is closed, then look for a contradiction.

I think you simply switched the work that has to be done and hid the essential parts behind "clear limit point". You have only shown stated, that the axis are boundary points. What is $n$ and is it sufficient that there is a neighborhood $N$?

Why don't you show, that every point $n_0=(x_0,y_0) \in U$ is the center of an open disc $B(n_0,r_0)$ that is completely inside $U$?
If it can be done positively, it is always better than a proof by contradiction. You could as well show, that $U^C$ is closed. This is already an equivalent definition of $U$ is open. So you don't need a contradiction. However you then have to show, that all limit points of $U^C$ are already in $U^C$. "Clear" means cheating.

 

[sa1988]

If $U^c$ is not open, it doesn't necessarily mean it's closed. In R, for example, the interval (0,1] is neither open nor closed.

I think a direct proof would be pretty straightforward. If $(x_0, y_0) \in U$, find an open set $N=(x_0-\epsilon,x_0+\epsilon)\times(y_0-\epsilon,y_0+\epsilon)$ where $\epsilon>0$ is appropriately chosen. It should be straightforward to show that every point in that set lies entirely in $U$.

I think you simply switched the work that has to be done and hid the essential parts behind "clear limit point". You have only shown stated, that the axis are boundary points. What is $n$ and is it sufficient that there is a neighborhood $N$?

Why don't you show, that every point $n_0=(x_0,y_0) \in U$ is the center of an open disc $B(n_0,r_0)$ that is completely inside $U$?
If it can be done positively, it is always better than a proof by contradiction. You could as well show, that $U^C$ is closed. This is already an equivalent definition of $U$ is open. So you don't need a contradiction. However you then have to show, that all limit points of $U^C$ are already in $U^C$. "Clear" means cheating.

Heh, funnily enough, in all my scrawlings trying to answer this, my first attempts used the methods both of you have suggested. I felt like I was getting to a dead end though.

From what I can understand, the neighbourhoods $N=(x_0-\epsilon,x_0+\epsilon)\times(y_0-\epsilon,y_0+\epsilon)$ and discs $B(n_0,r_0)$ both allude to the same concept, which is to find and show that every point in the intervals, or discs, is still in $U$.

But I don't really know how to show it. For a point $n \in N$, I don't know what I need to do to properly demonstrate that it's definitely in $U$ .

 

[vela]

You said $U$ consists of points in the first and third quadrants (excluding the axes). So you want to show that if you pick a point in the neighborhood, it's in one of those two quadrants. How would you mathematically describe the first and third quadrants?

(The reason I wrote N the way I did was because the problem referred to the product topology.)

 

[fresh_42]

You can measure in $\mathbb{R}^2$. So $(x_0,y_0)$ has a positive distance to the axis. Cut this distance in half - or a quarter to be safe - and take this as the diameter of your neighborhood (or the $\varepsilon$ in @vela 's correct notation where the disc (ball) is a cube.)

 

[sa1988]

You said $U$ consists of points in the first and third quadrants (excluding the axes). So you want to show that if you pick a point in the neighborhood, it's in one of those two quadrants. How would you mathematically describe the first and third quadrants?

The first quadrant is the points defined by the x and y intervals $\Big((-\infty, 0), (0,-\infty)\Big)$ and the third quadrant is the intervals $\Big((0,\infty), (0,\infty)\Big)$

But then how can I say with certainty that a point in any neighbourhood is within these limits, other than by saying that the neighbourhood is defined to be open so it can't ever go beyond those limits?

Would it be...

Take any point $(x,y)$ such that $xy > 0$. Then $(x,y)$ must be some value in the open intervals $\Big((-\infty, 0), (0,-\infty)\Big)$ or $\Big((0,\infty), (0,\infty)\Big)$, hence $U$ is open.

??

You can measure in $\mathbb{R}^2$. So $(x_0,y_0)$ has a positive distance to the axis. Cut this distance in half - or a quarter to be safe - and take this as the diameter of your neighborhood.

Going by your method with the open discs, would I be right in saying that the radius of these discs is always smaller than the distance to the axis?

The problem is that it all seems a bit circular to me, like I eventually just have to proclaim, "This is true because it was defined that way."

Foe example, I could take the open disc $B_\epsilon (x,y)$ with radius $\epsilon > 0$, but this can only work if $\epsilon$ is not so large that it pushes any value beyond the limits defined by $U$, and so it seems I have to again rely on the 'by definition' fallback.

 

[vela]

The first quadrant is the points defined by the x and y intervals $\Big((-\infty, 0), (0,-\infty)\Big)$ and the third quadrant is the intervals $\Big((0,\infty), (0,\infty)\Big)$

You got the first and third quadrants backwards. Anyway, a more useful way of saying a point is in, say, the first quadrant is to say x>0 and y>0. So can you show if a point is in N, that x>0 and y>0 (or x<0 and y<0 in the other case)?

Going by your method with the open discs, would I be right in saying that the radius of these discs is always smaller than the distance to the axis?

The problem is that it all seems a bit circular to me, like I eventually just have to proclaim, "This is true because it was defined that way."

Foe example, I could take the open disc $B_\epsilon (x,y)$ with radius $\epsilon > 0$, but this can only work if $\epsilon$ is not so large that it pushes any value beyond the limits defined by $U$, and so it seems I have to again rely on the 'by definition' fallback.

You want to show that you can define it that way. You can't do that for every point in a non-open set.

 

[fresh_42]

First of all, we may assume an arbitrary point $(x_0,y_0) \in U$ with $x_0 > 0$ and $y_0>0$, because the other case is symmetric.
Then you can define $\varepsilon := \min\{\frac{1}{2}x_0\, , \,\frac{1}{2}y_0\} > 0$.
So all you have to show now is, that every point of $N=(x_0-\varepsilon,x_0+\varepsilon)\times(y_0-\varepsilon,y_0+\varepsilon)$ is inside $U$, i.e. $N \subseteq U$. It's likely, that $N$ is in the first quadrant, so you have to show that an arbitrary point $(x,y)\in N$ satisfies $x > 0$ and $y>0$ with equality not allowed.

These are the definitions and the claim without any cheating. Now can you show this?

 

[sa1988]

You got the first and third quadrants backwards.

Ha, oops.

Anyway, a more useful way of saying a point is in, say, the first quadrant is to say x>0 and y>0. So can you show if a point is in N, that x>0 and y>0 (or x<0 and y<0 in the other case)?

For a point $n \in N$, it must have coordinates $(x,y)$ where $xy > 0$ by definition since $N \subset U$

This only true for $(x>0, y>0)$ or $(x<0, y<0)$, which are all open intervals.

Hence $N$ is open and therefore $U$ also open.

Is that ok??

First of all, we may assume an arbitrary point $(x_0,y_0) \in U$ with $x_0 > 0$ and $y_0>0$, because the other case is symmetric.
Then you can define $\varepsilon := \min\{\frac{1}{2}x_0\, , \,\frac{1}{2}y_0\} > 0$.
So all you have to show now is, that every point of $N=(x_0-\varepsilon,x_0+\varepsilon)\times(y_0-\varepsilon,y_0+\varepsilon)$ is inside $U$, i.e. $N \subseteq U$. It's likely, that $N$ is in the first quadrant, so you have to show that an arbitrary point $(x,y)\in N$ satisfies $x > 0$ and $y>0$ with equality not allowed.

These are the definitions and the claim without any cheating. Now can you show this?

Ahh, I think I get it now.

With $\varepsilon$ defined in that way, it follows that for $x, y > 0$ we have $x\pm \varepsilon >0$ and $y\pm \varepsilon > 0$

which means $(x\pm \varepsilon)(y \pm \varepsilon) > 0$

for any arbitrary $\varepsilon > 0$

(and the same can be applied when $x, y, < 0$)

which means $N$ is open, thus $U \supset N$ is open.

 

[fresh_42]

With $\varepsilon$ defined in that way, it follows that for $x, y > 0$ we have $x\pm \varepsilon >0$ and $y\pm \varepsilon > 0$

which means $(x\pm \varepsilon)(y \pm \varepsilon) > 0$

Almost. From $x>0$ you cannot conclude $x-\varepsilon > 0$. At least not:

for any arbitrary $\varepsilon > 0$

(and the same can be applied when $x, y, < 0$)

which means $N$ is open, thus $U \supset N$ is open.

Except that it's not for arbitrary $\varepsilon$. Only for small enough values. If you choose $(1,2)$ in $U$, then $(1-1000,1+1000) \times (2-1000,2+1000)$ is not in $U$. But we only have to find one open neighborhood (for each point).

For $(x_0,y_0) \in U$ and the $\varepsilon$ defined as above, we have $0 < \frac{1}{2}x_0 = x_0 - \frac{1}{2} x_0 \leq x_0 - \varepsilon$.
And similar for the second coordinate $y$ of your product topology. These are the two lines which build the bridge between the definitions (see post #9) and the fact there is an open $N \subseteq U\,.$ You should read again how the product topology is defined, in order to understand why @vela's definition of $N$ (see post #2) is the correct one and mine with the disc was a sloppy version of it.

 

[sa1988]

Almost. From $x>0$ you cannot conclude $x-\varepsilon > 0$. At least not:

Except that it's not for arbitrary $\varepsilon$. Only for small enough values. If you choose $(1,2)$ in $U$, then $(1-1000,1+1000) \times (2-1000,2+1000)$ is not in $U$. But we only have to find one open neighborhood (for each point).

For $(x_0,y_0) \in U$ and the $\varepsilon$ defined as above, we have $0 < \frac{1}{2}x_0 = x_0 - \frac{1}{2} x_0 \leq x_0 - \varepsilon$.
And similar for the second coordinate $y$ of your product topology. These are the two lines which build the bridge between the definitions (see post #9) and the fact there is an open $N \subseteq U\,.$ You should read again how the product topology is defined, in order to understand why @vela's definition of $N$ (see post #2) is the correct one and mine with the disc was a sloppy version of it.

Argh, sorry. Weirdly it seems that one of my downfalls in this area of maths is my lack of ability to choose the very specific words for making a logical point.

When I said 'arbitrary $\varepsilon$', it was meant as a hand-wavy nod towards the notion that $\varepsilon$ is essentially a tiny increment away from the point it's assigned to, and is always greater than 0, which makes the set open.

Yeah I figured the other definition of $N$ was more relevant to the question at hand, but I was working with your 'sloppy' definition because I've seen it used in proofs for other parts of my course, so it was still valuable to play around with.

Aside from the incorrect use of 'arbitrary', was I on the right track with the answer I came to?

 

[fresh_42]

Yes. And the difference between vela's definition and the disc is, that vela's neighborhood is according to the product topology whereas the disc neighborhood is according to the standard topology of $\mathbb{R}^2$ induced by the metric. Both topologies are equivalent, so it makes not really a difference, but they are not defined the same.

 

[sa1988]

Yes. And the difference between vela's definition and the disc is, that vela's neighborhood is according to the product topology whereas the disc neighborhood is according to the standard topology of $\mathbb{R}^2$ induced by the metric. Both topologies are equivalent, so it makes not really a difference, but they are not defined the same.

Great stuff, thanks.

To be honest, one thing confusing me is when a set is said to be 'equipped with a topology'.

What does it mean to be equipped with a topology? I understand that you can build a topology from a basis, and I understand that a topology only exists if it obeys the three main axioms, but I don't see how this allows for variation on what a set X can be 'equipped' with.

For example, if we have the set $X = \{2,4,6,8\}$, what does it literally mean to create a topology on $X$ ?

Currently I could say $\tau = \big\{\emptyset, \{2\}, \{6\}, \{2, 6\}\big\}$, and I believe I'm correct in saying that this is a topology on $X$.

Is it that a topology defines a specific space, where all the values in that space are fundamentally related by the fact that they're part of the original set $X$, along with any other rules specific to that topology?

For another more abstract example (and this is where I get more confused), I have another question on my exercise sheet which says,

"Consider the interval $[0,1]$ equipped with the subspace topology induced from the standard topology on $R$. "

Is this essentially a way of saying that we take all points in $[0,1]$ such that those points sit on open intervals in $R$, since this is how the standard topology is defined?

 

[fresh_42]

A topology on a set $X$ is simply the definition (or listing) which subsets (elements of $2^X=\mathcal{P}(X)$) are considered open.
As a set $U \subseteq X$ is open if and only if $U^C=X-U$ is closed, one can as well define all the closed sets.

Therefore "$X$ is equipped with a topology" means "defined which subsets of $X$ are open (or closed)".
It is important to notice, that open and closed are not opposites: there may be sets, which are neither open nor closed, e.g. $[0,1) \subseteq \mathbb{R}$, as well as sets, which are both, e.g. $\emptyset \, , \,X$. As to your example topology $\tau$: Yes, you can define whatever you want, as long as the three properties are satisfied. This means that you have forgotten to add $X=\{2,4,6,8\}$ as an open set to $\tau$.

The induced topology on a subset $Y \subseteq X$ is simpler as you think and I'm not sure I understand what you've said about $Y=[0,1]\subseteq \mathbb{R}=X$. All you have to do is, to define $V \subseteq Y$ as open if and only if there is an open set $U \subseteq X$, such that $V=U \cap Y$. This means for $Y=[0,1]$ we get as open sets all intervals $[0,a)\; , \;(a,1]\; , \;(a,b)$ with $0<a<b<1$ plus of course $\emptyset$ and $Y=[0,1]$ and all arbitrary unions and finite intersections of those.

 

[sa1988]

A topology on a set $X$ is simply the definition (or listing) which subsets (elements of $2^X=\mathcal{P}(X)$) are considered open.
As a set $U \subseteq X$ is open if and only if $U^C=X-U$ is closed, one can as well define all the closed sets.

Therefore "$X$ is equipped with a topology" means "defined which subsets of $X$ are open (or closed)".
It is important to notice, that open and closed are not opposites: there may be sets, which are neither open nor closed, e.g. $[0,1) \subseteq \mathbb{R}$, as well as sets, which are both, e.g. $\emptyset \, , \,X$. As to your example topology $\tau$: Yes, you can define whatever you want, as long as the three properties are satisfied. This means that you have forgotten to add $X=\{2,4,6,8\}$ as an open set to $\tau$.

The induced topology on a subset $Y \subseteq X$ is simpler as you think and I'm not sure I understand what you've said about $Y=[0,1]\subseteq \mathbb{R}=X$. All you have to do is, to define $V \subseteq Y$ as open if and only if there is an open set $U \subseteq X$, such that $V=U \cap Y$. This means for $Y=[0,1]$ we get as open sets all intervals $[0,a)\; , \;(a,1]\; , \;(a,b)$ with $0<a<b<1$ plus of course $\emptyset$ and $Y=[0,1]$ and all arbitrary unions and finite intersections of those.

Riiiight, I'm beginning to get it now, thanks. And yeah, I realize now that I missed $X=\{2,4,6,8\}$ for the topology I created, as it needs $X$ to satisfy the first axiom, heh.

So now what I'm generally getting from this is that the topology on a set is essentially a way of setting out the first step in how the elements in that set are to be used or understood for the given problem.

From this, I think I'd be right in saying that metrics can (or must?) then be applied to really start making use of what topology is about? For example, if you place the standard topology on $R^2$ and use a metric $d(x,y) = max|x_i - y_i|$, you can create an open square with the balls $B_\epsilon (x) = \{y \in X : d(x,y) < \epsilon \}$ , so the square has a vertical and horizontal radius $\epsilon$. And this only works because $R^2$ is equipped with the standard topology which means all the elements in that square are open, hence a square is described.

Am I on the right track with my understanding?

Sorry for taking the thread off topic and throwing my own musings out. I'm just trying to get a grip on the wider idea of topology itself and what it's about. So far it's the most 'non-physics' feeling of a subject I've done so far, because it's all just pure maths at the moment, so I need to quickly get to grips with what's going on.

 

[fresh_42]

So far it's the most 'non-physics' feeling of a subject I've done so far, because it's all just pure maths at the moment, so I need to quickly get to grips with what's going on.

O.k. "so far" is probably right, since you certainly know it better than me. In general it's not true. Relativity theory, particle physics, differential equations, esp. because of the initial conditions, and thus basically everything in physics involves topologies. There is a fundamental difference between open and closed sets and important to get a feeling about them. Mostly you will meet topologies induced by a metric, because in the end, physics is about measurements and therefore distances in some context. Also important here is, that mathematical objects come in pairs: objects and mappings between those objects. E.g. vector spaces and linear mappings, manifolds and diffeomorphisms, groups and homomorphisms, functions and operators, and in this case, topolopgies and continuous functions. You need a topology to define continuity, because a function is continuous in $\xi$ if for every open neighborhood $U$ of $f(\xi)$ the pre-image $f^{-1}(U)$ is an open neighborhood of $\xi$. This shows the local nature of the concept and that it is not about $\xi$ but the neighborhoods of $\xi$ and $f(\xi)$. It also shows the physical importance, since really many functions, which are investigated in physics are continuous.

Riiiight, I'm beginning to get it now, thanks. And yeah, I realize now that I missed $X=\{2,4,6,8\}$ for the topology I created, as it needs $X$ to satisfy the first axiom, heh.

Yes.

So now what I'm generally getting from this is that the topology on a set is essentially a way of setting out the first step in how the elements in that set are to be used or understood for the given problem.

Here you begin to miss the essential point. Open and closed are local properties of a space. It has nothing to do with the elements in the space. It is about its regions, i.e. its subsets. Intuitively it's simply: without the boundary then open, with the boundary then closed. Of course this doesn't mean a lot if you consider finite sets, but usually we deal with infinite sets like functions with some property, the Euclidean space or various other spaces. As you see, the elements or their nature are not decisive here, neighborhoods and (infinite) sequences of points are.

From this, I think I'd be right in saying that metrics can (or must?) then be applied to really start making use of what topology is about? For example, if you place the standard topology on $R^2$ and use a metric $d(x,y) = max|x_i - y_i|$, you can create an open square with the balls $B_\epsilon (x) = \{y \in X : d(x,y) < \epsilon \}$ , so the square has a vertical and horizontal radius $\epsilon$. And this only works because $R^2$ is equipped with the standard topology which means all the elements in that square are open, hence a square is described.

The standard topology usually refers to the Euclidean metric $d(x,y)=\sqrt{x^2+y^2}$. Because of this we talk about open balls. Didn't it sound strange to you as you wrote "the square has a vertical and horizontal radius"? But you are right as it isn't important which metric of the two we chose here. It won't change the properties of the topological space $\mathbb{R}^2$. But "ball" and "diameter" is a bit weird, if your basis open sets are cubes. A metric is nothing else than a ruler to measure distances. O.k. with spaces of functions or sequences, the ruler might be not as obvious as if we measure the Euclidean distance, but in principle it's nothing else. Now what does "by the metric induced topology" mean? As you wrote, we take the balls $B_\varepsilon(x) = \{y \in X \, : \, d(x,y) < \varepsilon\}$ as open sets. With them we have infinite unions of them as open sets as well. And again, since we define those open balls by a strictly $<$ relation, the boundaries don't belong to them. If we include the boundaries, i.e. $d(x,y) \leq \varepsilon$, then we get closed sets $\overline{B}_\varepsilon(x)$. And if you remove some (not all) points, for which $d(x,y)=\varepsilon$ holds, then this set is neither open nor closed. So you see, it's not about points, it's all about what is around them. And with a metric, we have a measure to define "around". Without a measure (metric), we have to define what we mean by "around", e.g. via a list of open sets if nothing else is available.

 

[sa1988]

Right, so a topology is basically a collection of open sets taken from a given set - e.g. the standard topology on $R^2$ is the union of all open balls $B_r (x)$ of radius $r$ about each and every point $x$, which is infinite.

... And then a metric defines how to interpret the difference between two points - e.g. the Euclidian metric tells you that $x$ and $y$ can be compared by taking $d(x,y) = \sqrt{x^2 + y^2}$. Without that metric, the points have almost no meaning.

... And then a function is what manipulates those points so as to 'create' a thing that wants to be looked at - e.g taking a topology generated by the basis $\beta_\epsilon = \{y \in X : d(x,y) < \epsilon \}$ and employing the function $f: R^2 \rightarrow R^2$ which maps $(x,y) \mapsto (2 sin(x), 2 sin(y))$, would be a representation of an open circle of radius 2.

So the topology is the the overall 'thing' being worked with and is how you define what is open or closed from the start; the metric is what allows comparison between points in the topology, and a function essentially creates 'objects' using the topology. And this is where homeomorphisms come in, for example the classic "donut = teacup" line, where presumably the only difference is in the function being employed.

How's my understanding now? Any gaping holes?

Thanks for this by the way - gigantically appreciated.