# Contents

## Idea

A continuum is in general something opposite to a discrete. There are several notions of continuum in mathematics:

In physics one may also mean by a continuum a medium which spreads the physical quantities spatially with some finite density, unlike the physics of a system of particles, where an infinite (delta-distribution-like) density is attached to a discrete system of points. See at geometry of physics for more on the relation of the continuum in form of the real line to physics.

## In cohesive homotopy type theory

One can axiomatize aspects of the notion of line continuum in cohesive homotopy type theory. There the idea of an object $\mathbb{A}^1$ all whose points are, while different, connectable by continuous paths (and uniquely so, up to suitable homotopy) is encoded in asking that after applying the fundamental ∞-groupoid functor $\mathbf{\Pi}$ to it, the result is something contractible

$\mathbf{\Pi}(\mathbb{A}^1) \simeq * \,.$
###### Example

In the model of cohesive homotopy type theory called Smooth∞Grpd we have a full and faithful embedding of smooth manifolds. Therefore we can embed the integers $\mathbb{Z}$, the rational numbers $\mathbb{Q}$ as well as the real numbers $\mathbb{R}$, all equipped with their canonical smooth manifold structure. This is discrete for the first two, but not for the last one, and homotopy cohesion can detect this:

$\mathbf{\Pi}(\mathbb{Z}) \simeq \mathbb{Z} \,;$
$\mathbf{\Pi}(\mathbb{Q}) \simeq \mathbb{Q} \,;$

but

$\mathbf{\Pi}(\mathbb{R}) \simeq * \,.$

This reflects the fact that the points of $\mathbb{R}$ form a continuum, but those of $\mathbb{Z}$ and $\mathbb{Q}$ do not.

Also the complex numbers $\mathbb{C}$ with their canonical manifold structure of course form a continuum in this sense

$\mathbf{\Pi}(\mathbb{C}) \simeq * \,.$
###### Proposition

For a ring object $\mathbb{A}^1$ to be geometrically contractible, $\Pi(\mathbb{A}^1) \simeq *$, it is sufficient that there be a map $i$ from a bipointed type $(left, right, I)$ to the bipointed type $(0, 1, \mathbb{A}^1)$ such that $I$ is geometrically contractible.

Hence in words: “If in the ring $\mathbb{A}^1$ the elements 0 and 1 are path-connected, then $\mathbb{A}^1$ is already contractible, hence is a line continuum.”

###### Proof

Under the given assumptions we obtain a commuting diagram of the form

$\array{ \mathbb{A} \\ \downarrow & \searrow^{\mathrlap{ 0}} \\ I \times \mathbb{A}^1 &\stackrel{ mult\circ (i,id)}{\to}& \mathbb{A}^1 \\ \uparrow & \nearrow_{\mathrlap{id}} \\ \mathbb{A}^1 } \,.$

Since $\Pi$ preserves products, it sends this to a diagram of the form

$\array{ & \nearrow \searrow^{\mathrlap{0}} \\ \Pi(\mathbb{A}^1) &\Downarrow_{\simeq}& \Pi(\mathbb{A}^1) \\ & \searrow \nearrow_{\mathrlap{id}} } \,,$

which exhibits a contracting homotopy of $\Pi(\mathbb{A}^1)$.

###### Example

The standard unit intervals $[0,1] \hookrightarrow \mathbb{R} \in$ TopMfd $\hookrightarrow$ ETop∞Grpd and $[0,1] \hookrightarrow \mathbb{R} \in$ SmthMfd $\hookrightarrow$ Smooth∞Grpd satisfy the assumptions of prop. 1.

###### Remark

The $I$ of prop. 1 is in general not an interval type, but only its image $\Pi(I)$ is. See the discussion Geometric spaces and their cohesive homotopy types at cohesive homotopy type theory for more on this.

## References

Revised on September 27, 2013 12:36:32 by Urs Schreiber (158.109.1.23)