# Tag Info

28

If the person is in a Möbius strip, then it seems we are assuming he is $2$-dimensional. Suppose he has with him two identical circles split into sectors of $120^{\circ}$, and each sector is colored a different color. Notice being $2$-dimensional, he can rotate this circle but not reflect it, so the two circles are identical up to a rotation. Now, let him ...

20

Yours is a very interesting and subtle question, which often generates confusion. First let us give a name to the property you are interested in: a ring $A$ will be said to satisfy (DIM) if for all $\mathfrak p \in \operatorname{Spec}(A)$ we have $$\operatorname{height}(\mathfrak p) +\dim A/\mathfrak p=\dim(A) \quad \quad (\text{DIM})$$ The main ...

15

Quoting from Abbot's book, "Understanding Analysis": There is a sensible agreement that a point has dimension zero, a line segment has dimension one, a square has dimension two, and a cube has dimension three. Without attempting a formal deﬁnition of dimension (of which there are several), we can nevertheless get a sense of how one might be deﬁned by ...

10

Your ideal is generated by binomials, so one can be smart about it. There is an algebra map $\mathbb C[x_1,x_2,x_3,x_4]\to \mathbb C[s, t]$ such that $x_1\mapsto s^3$, $x_2\mapsto s^2t$, $x_3\mapsto st^2$ and $x_4\mapsto t^3$, and this map maps your ideal $\mathfrak p$ to zero, so it induces $\phi:\mathbb C[x_1,x_2,x_3,x_4]/\mathfrak p\to \mathbb C[s, t]$. ...

9

Negative dimension is actually much easier to talk about than complex dimension. Super vector spaces are a natural collection of objects that can have negative dimension; given a super vector space $(V_0, V_1)$ we can define its dimension to be $\dim V_0 - \dim V_1$, and this definition has many nice properties; see this blog post, for example. More ...

9

The most basic surprise, in my opinion, is that the ratio of the volume of the unit sphere to the volume of the cube circumscribing that sphere tends to 0 as the dimension of the space tends to $\infty$. In other words, a high-dimensional sphere takes up almost no space in the cube that circumscribes it. See pp.4--5 in ...

9

I don't know if this is on Cover's list, but maybe it should be: For $n=2$ and $3$, any tiling of ${\mathbb R}^n$ by unit $n$-cubes has two with a complete facet in common. But it's not true for $n \ge 10$: see http://arxiv.org/pdf/math.MG/9210222.pdf

8

suppose $P_0, \ldots, P_n$ are $n+1$ polynomials, of degree less than $d$. Then by multiplying the $P_i$ among themselves up to $k$ times, you can build at least about $k^{n+1}/(n+1)!$ polynomials of the form $\prod P_i^{\alpha_i}$ of degree less than $dk$. But the dimension of the vector space of polynomials of degree less than $dk$ in $K[X_1,\ldots X_n]$ ...

8

There are formal ways to define dimension, indeed, lots of them, depending on the context. As already noted in a comment, your description of why a circle is one-dimensional is not really correct though. The one-dimensionality of a circle is not a function of the fact that a circle itself can be described by a single number (such as radius), but that to ...

7

This should never be true for a reasonable definition of dimension (for example the dimension of a manifold). A lower-dimensional thing should have measure zero in a higher-dimensional thing, so removing it shouldn't change the dimension of the higher-dimensional thing. The correct version of the "naive equation" is that the Cartesian product of an ...

7

Standard self-similarity Fractals are often constructed using a recursive procedure and self-similar sets in particular, are always constructed this way. I think that most folks working in fractal geometry would guess that your picture implies a recursive construction like so: Note that the first set is an initial seed. The second set is composed of ...

7

What is the rule or condition to be a 2D or 3D picture? This is a subtle question! In a certain sense, the picture is two-dimensional, since it's displayed on a computer screen. :) That's a trivially literal remark, yet not without mathematical content. A closely related question is, "Does there exist a physical object that looks like the picture from ...

6

There are many generalizations of the usual notion of dimension, and they are there to capture different properties. Having said that, the intuition behind the dimension is that it describes the number of degrees of freedom you have, e.g. A point on a line has one degree of freedom. A point on a plane has two degrees of freedom. A point of two-dimensional ...

6

It is possible to force this perspective, but it would be incorrect to say that the mathematics "actually" is this way. To be more specific, yes, you can choose to consider the inclusions $j_n\colon \mathbb{R}^n\longrightarrow\mathbb{R}^\infty$ defined by $$j_n(x_1,\ldots,x_n)=(x_1,\ldots,x_n,0,0\ldots)$$ However, the objects $\mathbb{R}^n$ have their own ...

5

Hausdorff outer measure is defined for all sets, and then we use the definition of Caratheodory to restrict it to a subalgebra of "measurable" sets to get the Hausdorff measure. In $\mathbb R^n$, the $n$-dimensional Hausdorff outer measure is the same (up to a constant factor) as $n$-dimensional Lebesgue outer measure, so they have the same measurable sets ...

5

Here, adapted from an example and a problem in Engelking and with lots of blanks filled in, is an example of a zero-dimensional Tikhonov space with a subspace $-$ in fact a closed subspace $-$ of dimension greater than $0$. The first step is to construct a zero-dimensional Tikhonov space $X$ that is not strongly zero-dimensional; this construction is ...

5

For one, it gives an invariant which helps us distinguish sets which are otherwise rather difficult to tell apart, and in a way which is reflected in actual, interesting properties of the sets. On the other hand, there are many situations in which the Hasdorff dimension of a set controls analytical properties of solutions of certain differential equations ...

5

You are correct. There are several ways to show that it is impossible, and it basically boils down to the fact that no open subset of ${\bf R}^n$ is homeomorphic to an open subset of ${\bf R}^m$ if $n<m$ (as $U\cap V$ would be in your case). You can assume without loss of generality that the sets in question are connected. One way to show it is to use ...

5

$\omega^\omega$ can be visualized, in what I think is a fairly nice way in a static 2D image featured on the wikipedia page for ordinal number: Also, if you're willing to allow dynamic visualizations, then Stephen Brooks's transfinite number line goes well past $\epsilon_0$ (to $\Gamma_0$), as well as providing a more linear (if colorful) look at ...

5

I think Alexander Duality is what you are looking for. I gather that you are a non-expert, so I will attempt to describe in fairly informal terms how Alexander duality deals with the questions that you are interested in. Consequently, I'll suppress the inevitable technicalities, since they don't enter into the very geometric situations that you are ...

4

Concentration of measure phenomena provide great examples of how our intuition based on low-dimensional space is unreliable in high-dimensions. Compare unit balls in the metric spaces $\mathbb R^n$ endowed with, resp. the Euclidean metric $L_2$, versus $L_1$, and $L_{\infty}$. The unit balls of $L_2$ are bounded by "round" spheres and are sandwiched ...

4

We expect a normally-distributed random variable to take values close to the mean, and in low dimensions it does. But in high dimensions, it does not. The volume of a thin hyperspherical shell increases so rapidly as its radius increases that even though the variable has greatest probability density near the mean, most of the probability mass is far from ...

4

If he has a friend then they both can paint their right hands blue and left hands red. His friend stays where he is, he goes once around the strip, now his left hand and right hand are switched when he compares them to his friends hands.

4

Choose a basis for $W_1$. This basis can be extended to a basis for $V$ (well-known result). Now let $W_2$ be the span of the newly added vectors. Then $W_1 \cap W_2 = \{0\}$ by linear independence and $W_1 + W_2 = V$.

4

These definitions are not the same in general, if $M$ is not f.g. Consider the module $\mathbb Q_p/\mathbb Z_p$ over $\mathbb Z_p$. Its annihilator is $0$, so the first definition gives dimension $1$. On the other hand, its support is the closed point of Spec $\mathbb Z_p$, and so the second definition gives dimension $0$. If you are reading an article ...

4

Let's stick to separable metric spaces; without separability the Hausdorff dimension is always infinite. The following theorem can be found in Dimension Theory by Hurewicz and Wallman: $\inf\{\dim_{\mathcal H} Y : Y \text{ homeomorphic to } X \} = \dim_{\mathcal T} X$ where $\dim_{\mathcal T}X$ is the topological dimension of $X$. Hence, the spaces ...

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