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Are planes in $3$-dimensions two-dimensional?

The reason I ask is because mathematically the $xy$-plane exists in $3$D space but appears to be $2$D, but how can something $2$D be in $3$D space? I therefore conclude that a plane in $3$D space must have a thickness that is infinitesimal. Isn't this a contradiction, though, since planes have a thickness of that of a point and since a point has no thickness neither does the plane?

user19405892
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2 Answers2

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The first thing to mention when talking about the term dimension is that in mathematics there is an exact definition of this term. So, when we speak about dimensions, the first thing to do is to define what dimension means. For the kind of geometry you use, the definition of the term dimension from a mathematical field called linear algebra is the appropriate one. In this sense, dimension is a property of so-called vector spaces.

  • Let's start by defining the term vector space (I will only consider vector spaces over $\mathbb{R}$ here to simplify things): The set

    $V=\{(x_1,x_2,\ldots,x_n)\mid x_1,x_2,\ldots,x_n\in\mathbb{R}\}$

    is a vector space of dimension $n$. Please note: Not every vector space is of this form!

  • Remark: From school you may know $\mathbb{R}^2$ (which is a vector space of dimension 2) and $\mathbb{R}^3$ (which is a vector space of dimension 3).

  • The interesting question is: Given a subset $U$ of $V$, what is its dimension? We can only define what the dimension of $U$ is if $U$ itself is a vector space. $U\subseteq V$ is a vector space, if and only if there exist $v_1,v_2,\ldots,v_m\in V$ such that

    $U=\{a_1v_1+a_2v_2+\ldots+a_mv_m\mid a_1,a_2,\ldots,a_m\in\mathbb{R}\}$

  • Now two cases can occur: Either $U=V$ holds or $U\subset V$ holds. For $U=V$, the dimension of $U$ is obviously $n$. For $U\subset V$, we still have to define what dimension means.

  • For defining the term dimension, we first have to define what linearly independent means: A set $W=\{w_1,w_2,\ldots,w_m\},W\subseteq V$ is called linearly independent, if the equation

    $a_1w_1+a_2w_2+\ldots+a_mw_m=\mathbf{0}\quad(a_i\in\mathbb{R})$

    only holds for $a_1=a_2=\ldots=a_m=0$.

  • For example: The set $\{(1,0,0),(0,1,0),(3,2,0)\}$ is not linearly independent, because

    $3\cdot (1,0,0)+2\cdot (0,1,0)-(3,2,0)=(0,0,0)$

    On the other hand, the set $\{(1,0,0),(0,1,0)\}$ is linearly independent, because

    $a_1\cdot (1,0,0)+a_2\cdot (0,1,0)=(0,0,0)$ implies $a_1=a_2=0$.

  • Now we can define what dimension means:

    $U=\{a_1v_1+a_2v_2+\ldots+a_mv_m\mid a_1,a_2,\ldots,a_m\in\mathbb{R}\}$

    has dimension $m$ if and only if $\{v_1,v_2,\ldots,v_m\}$ is a linearly independent set.

  • Example (continuation of the above example):

    $\{a_1(1,0,0)+a_2(0,1,0)+a_3,(3,2,0)\mid a_1,a_2,a_3\in\mathbb{R}\}$

    is the same vector space as

    $\{a_1(1,0,0)+a_2(0,1,0)\mid a_1,a_2\in\mathbb{R}\}$;

    its dimension is 2 (because $\{(1,0,0),(0,1,0)\}$ is linearly independent).

  • Remark: Given a vector space $U$, its dimension is unique.
  • Remark: If $n$ is the dimension of $V$, $m$ is the dimension of $U$ and $U\subseteq V$, then $m\leq n$ holds.

  • If $U\subseteq V$ is a vector space and $v_0\in V$,

    $U^\prime=v_0+U=\{v_0+a_1v_1+a_2v_2+\ldots+a_mv_m\mid a_1,a_2,\ldots,a_m\in\mathbb{R}\}$

    is called an affine vector space. The dimension of $U'$ is defined as the dimension of $U$.

In the above points, we have defined the term dimension for vector spaces. What remains to be done is to show how vector spaces are related to basic geometry:

  • Any point can be represented by an affine vector space $\{v_0\}\subseteq V$. Therefore, points have dimension 0.
  • Any line can be represented by an affine vector space $\{v_0+a_1v_1\mid a_1\in\mathbb{R}\}$, where $v_0\in V$ and $v_1\in V$ are given. Therefore, lines have dimension 1. (The set contains the points that lie on the line.)
  • Any plane can be represented by an affine vector space $\{v_0+a_1v_1+a_2v_2\mid a_1,a_2\in\mathbb{R}\}$, where $v_0\in V$, $v_1\in V$ and $v_2\in V$ are given and $v_1,v_2$ are linearly independent. (The set contains the points that lie in the plane.)

Please note: There is no limit on how big a dimension can be; for example, you can also define vector spaces of dimension 1000.

Xaver
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Yes, it is two dimensional. We define the dimension of a vector space (or a subspace of a vector space) to be the cardinality of a minimum spanning set. Any plane in $\mathbb{R}^3$ can be written as the span of two vectors.

Stella Biderman
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