Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was reading about dimensions and in the Wikipedia article it states the following:

In mathematics, the dimension of an object is an intrinsic property independent of the space in which the object is embedded. For example, a point on the unit circle in the plane can be specified by two Cartesian coordinates, but a single polar coordinate (the angle) would be sufficient, so the circle is $1$-dimensional even though it exists in the $2$-dimensional plane. This intrinsic notion of dimension is one of the chief ways the mathematical notion of dimension differs from its common usages."

Taken from Dimension article at Wikipedia.

How could I know if a square is a $2$D object (which I'm sure it probably is), or that a circle is $1$D (which previously I believed it to be $2$D)?

Is there a way for any object to tell what dimensional object it is?

I don't really know how to tag this question, sorry.

share|cite|improve this question
3  
You are probably thinking of a disc. A disc is a circle including the area inside the boundary. A circle is just the curved line which forms the boundary of the disk. The circle is 1D while the disk is 2D. Another way to think of dimension is how many degrees of freedom you have in your movement if you were stuck in a circle or a disk. If I was confined to living on a circle, then at any point on the circle, I have only one degree of freedom. I can move "left" or "right". But if I was living in a disk, I can move left/right or I can move up/down so I have two degrees of freedom. – Fixed Point Mar 13 at 0:57
    
Ok, that makes more intuitive sense – frog1944 Mar 13 at 1:21
    
As a further complication, we have objects called "fractals" that feature a fractional dimension; the Koch snowflake has a dimension of $\approx 1.26186$, for instance. – J. M. Mar 13 at 10:39
    
So you can have non integer dimensions? Can you have negative dimensions @J.M. ? – frog1944 Mar 13 at 10:43
    
@frog1944, the framework I based my answer on is called differential geometry (this is the branch of mathematics most closely allied to mathematical physics) and in this framework, dimension is always a natural number. However, there's other frameworks and/or viewpoints on dimension (though less closely allied to physics). One of these frameworks is called measure theory. From this viewpoint, subsets of $\mathbb{R}^n$ can have "dimension" anywhere in the real interval $[0,n].$ Except that there's many non-equivalent... – goblin Mar 13 at 12:55
up vote 32 down vote accepted

The concept of dimension is surprisingly subtle. The mathematics invented by Georg Cantor and his contemporaries famously showed that, contrary to intuition, it is possible to specify a point in $2$-dimensional space using only a single real number. To make this precise, we need the concept of a bijective function. What Cantor's mathematics shows is that, contrary to intuition, there exist (many) bijections $$\mathbb{R} \rightarrow \mathbb{R} \times \mathbb{R}.$$

In some sense, this is saying that the set $\mathbb{R} \times \mathbb{R}$ is (rather paradoxically) "no bigger" than $\mathbb{R}$.

To make matters worse, Giuseppe Peano (born 13 years after Cantor) managed to construct a space-filling curve; a continuous function $$[0,1] \rightarrow [0,1] \times [0,1]$$ that, rather miraculously, manages to be surjective.

What this all means is that, like I said, the concept of dimension is rather subtle. One might speculate that fundamentally, this concept actually makes no sense. The good news is that, in fact, the concept of dimension does make sense. The bad news is that the definition is pretty complicated, and needs to be built up in two stages.

In the first stage, we establish the concept of dimension in linear algebra.

We need the following notions:

The definition is:

Proposition 0. Let $V$ denote a vector space. Then $V$ has a basis, and every two bases of $V$ have the same number of elements, and (Definition.) we call this number the dimension of $V$.

Example. The dimension of the vector space $\mathbb{R}^n$ is $n$.

Note that, in full generality, the dimension of $V$ is a cardinal number, a concept invented by Cantor to tame the chaos of infinite sets. However, for the purposes of basic geometry, we can usually assume that $V$ is finite-dimensional, in which case the dimension of $V$ will always be a natural number.

In the second stage, we establish the concept of dimension in differential geometry. This part is much more complicated.

We'll need the following concepts:

  • manifold (sometimes called a smooth manifold, or equivalently, a differentiable manifold).
  • tangent space

Definition. Let $M$ denote a smooth manifold and $x$ denote an element of $M$. Then the dimension of $M$ at $x$ is, by definition, the dimension of the tangent space of $M$ at $x$, viewed as a vector space.

It turns out that

Proposition 1. Let $M$ denote a manifold. If $M$ is connected, then there exists a natural number $n$ such that for all points $x$ in $M$, the dimension of $M$ at $x$ equals $n$, and (Definition.) we call $n$ the dimension of $M$.

Example. The dimension of the manifold $\mathbb{R}^n$ is $n$.

To finally answer your question, we'll need two more concepts:

Now:

  • the open interval $A=(0,1)$ can be viewed as a submanifold of $\mathbb{R}$
  • the open square $B=(0,1) \times (0,1)$ can be viewed as a submanifold of $\mathbb{R}^2$
  • the unit circle $C = \{x \in \mathbb{R}^2 : \|x\|=1\}$ can viewed as a submanifold of $\mathbb{R}^2$.

This implies that each of $A,B$ and $C$ can be viewed as manifolds in their own right. Further to this, they turn out to be connected; and, hence, they have a well-defined dimension; namely, $1,2$ and $1$ (respectively). The first two of these numbers is easy to obtain; since $A$ and $B$ are open subsets of $\mathbb{R}$ and $\mathbb{R}^2$ respectively, hence they have dimension $1$ and $2$ respectively. The dimension of $C$ is a little harder to find, because its not an open subset of $\mathbb{R}^2$. But your intuition is generally pretty trustworthy when it comes to these things; if your brain tells you that the dimension of $C$ is $1$, then the dimension of $C$ is probably $1$. Perhaps this partly explains why it took mathematics so long to give the concept of dimension a proper and rigorous treatment; perhaps its because intuition alone is usually enough to get you the right answer, even if you don't really know what that answer means in a precise, technical sense.

share|cite|improve this answer
3  
Thank you so much for your reply!!! It's so insightful and concise! I'd love to learn more about this sort of field of mathematics, any idea where I should start? Is there a name to this field of mathematics? – frog1944 Mar 12 at 7:07
2  
@frog1944 Start with Linear algebra, then go into differential geometry or topology (Though to start with the very underlying stuff at the beginning, that's set theory) – Alan Mar 12 at 9:00
    
great answer! But could you provide an intuitive explanation of why a square has dimension 2 while a circle has dimension 1? – Ant Mar 12 at 9:26
1  
@Ant: A filled-in square is two dimensional (and so is a filled-in circle). The borders of a square or circle are one dimensional. (Think of how many directions you can move if you're "standing on" them.) – Deusovi Mar 12 at 9:29
1  
A bit of feedback, if I may: the question seems to be getting at some kind of procedure that you can apply to a set of points (though obviously not an arbitrary set) to produce the set's dimension. This answer talks about representing points in, say, $(0,1)^2$ as a vector space which turns out to have two dimensions. But one might then wonder why you can't do the same thing with a space-filling curve, thereby representing $(0,1)^2$ as a one-dimensional vector space. It feels a little incomplete not to address that. (Granted, that might be more than an answer's worth.) – David Z Mar 12 at 20:35

While goblin's answer is right, its a little specific. A more common notion is covering dimension, which is both applies in more situations, and is a little easier to understand.

First we start with the notion of open set. An open set is one that does not contain any points in its boundary. For example, the set of points less than $1$ away from $(0,0)$ is open, because the boundary is the set of points exactly $1$ away from $(0,0)$, and that set is disjoint from the first one. The set of points less than or equal to $1$ away from $(0,0)$ is not open, since it contains some points, all of the points actually, that are in its boundary.

Now, if you take a line and cover it with blobs that are open sets, there will be some points covered by $2$ blobs. Since open sets don't contain their boundaries, if you try putting them right next to each other, they either have to overlap or have space in between them. If you do the same with the plane, some points will be covered by $3$ blobs. If you the same with space, some will be covered by $4$, etc...

So basically, the dimension of space is the max number of blobs to cover a point minus $1$.

We need to take one more thing into account though. Take a circle. If you cover it with tiny blobs, some of its points will have to be covered by $2$ blobs, so it is $1$ dimensional. But also notice that you could cover the entire circle with one big blob, suggesting it is $0$ dimensional. For this reason, we require the blobs to be arbitrarily small.

But what if we don't have a notion of size in our space? That's fine. What we say is that a space is dimension $d$, if when you cover the entire thing with blobs that are open sets, you can shrink (replace with a subset that is also an open set) the open sets such that each point is covered by at most $d+1$ blobs.

Notice how all this required was a notion of boundaries of sets. We didn't need to define distance or anything else, just boundaries.

Also note that if we viewed the circle as a space in and of itself, this still works. The boundaries of a set of points on the circle is just the set of any endpoints it has. The circle's dimension will be the same whether considered as part of the plane or as a space unto itself.

See https://en.wikipedia.org/wiki/Lebesgue_covering_dimension for more details.

share|cite|improve this answer
    
Thank you @PyRulez – frog1944 Mar 12 at 21:00
2  
Interesting! I didn't know any of this. – goblin Mar 13 at 13:00
    
very nice! Could this be somehow generalized to fractals with non-integers dimension? – Ant Mar 13 at 15:28
    
@Ant As it turns out, something like say, Koch's snowflake, is topologically equivalent to a circle. You need to at least have a notion of distance before you can define fractal dimension. – PyRulez Mar 13 at 15:36

A square is actually 1 dimensional, as you can use a single parameter to move around the edges of the square. Basically, the dimension is the minimum amount of parameters you need to describe a point in the space. More complicated, you need to go into the definition of manifolds and local diffeomorphisms

share|cite|improve this answer
    
Ok, so for any object is there a way to find out what dimension it is in (like an equation of sorts)? Or does that get complicated and require manifolds like you said in your answer. Furthermore how do we know a true nD object is really nD (where n doesn't equal 1)? If it is the minimum amount of parameters then how do we know that if we haven't "discovered" a new way of representing it, it would become a lower dimensional object. e.g. If we hadn't found out that there is such thing as an angle, we couldn't express the circle in 1D (I believe). – frog1944 Mar 12 at 5:42
    
The way you know is by establishing local difffeomorphisms between your space and k-dimensional Euclidean space. There is only one such k that such a function is possible for. Intuitively, you can look at it and go "is it only points", then it's dimension 0. If it's a line or circle or other thing you can squeeze and deform, to a line, then it's 1. If it's something you can squish around without tearing a circle or square to get (like a cylander, by pasting the sides of a rectangle), it's 2. Etc. – Alan Mar 12 at 6:02
    
The area of study for this starts with Multivariable Analysis/General Topology, then you can go into Algebraic topology. – Alan Mar 12 at 6:03
1  
Thank you very much @Alan :) – frog1944 Mar 12 at 7:08
1  
A square is also homotopic to $S^1\times [0,1]^{405}$, so squares are actually 406-dimensional o.O – Eric Stucky Mar 12 at 23:48

In topology there are 3 main types of dimension: small inductive dimension (ind), large inductive dimension (Ind), and covering dimension. As the article says, these are intrinsic (internal) properties of a space, not affected by any space that it is embedded in. There is also the Hausdorff-Besicovitch dimension, which does depend on the embedding space, and can have fractional values. In algebra,the word also has other meanings.

In a topological space $S,$ a neighborhood-base for a point $p\in S$ is a family $F$ of open sets, each containing $p,$ such that for any nbhd $U$ of $p,$ there exists $f\in F$ with $f\subset U.$

The small inductive dimension ind$(S)$ is $0$ iff (1)$S=\phi$ or (2) every $p\in S$ has a nbhd-base $F$ such that $\forall f\in F\;(\partial f=\phi)$.

And ind$(S)=n+1$ iff (1)$\; \neg ($ ind$(S)\leq n)$ and (2) every $p\in S$ has a nbhd-base $F$ such that $\forall f\in F\;($ind $(\partial f )\leq n).$

And if $\neg ($ind$(S)=n)$ for every $n\in N$ then ind$(S)=\infty.$

Note: $\partial f=\bar f \cap \overline {S\backslash f}$.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.