How to tell what dimension an object is? 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.
 A: 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
A: 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:  $\partial f=\bar f \cap \overline {S\backslash f}$.
A: 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:


*

*vector space

*basis
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:


*

*submanifold

*open set
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.
A: 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.


