show/hide this revision's text 3 Corrected the definition of S^n

As noted in the comment of Ryan Budney to Richard Kent answer, the non-compact surfaces have been classified (in a very nice way, hence my need to share the details).

I have no access to Jstor right now, so I rely on my memory and, well, you may want to check all this.

Ian Richards theorem says that non-compact surfaces (without boundary) are classified by their orientablility, their genus (possibly infinite) and a triple of spaces, each one embedded in the preceding, that are:

  1. the space of its ends,

  2. the space of its ends with genus,

  3. the space of its unorientable ends.

The space of ends is constructed by taking an increasing sequence of compact subset that cover a topological space $T$, and looking at the connected components of their complements. An end of $T$ is an infinite, decreasing sequence of such connected components. The point is that you can do that in a way that does not substancially depend on the sequence of compact you chose.

For example, $\mathbb{R}^n$ has only one end provided $n\ge2$ (look at the sequence of balls of integer radius and centered at some point), while $\mathbb{R}$ has two ends. The space of ends of a regular tree is a Cantor set.

An end is said to have genus if the connected components that define it all have genus (they never reduce to annuli). An end is said to be unorientable if the connected components that define it all are unorientable.

Now, consider the surface $S^n$ ($n=1,2$ or $3$ defined as the boundary of a tubular neighborhood of the usual embedding of the usual Cayley graph of $\mathbb{Z}^n$ into $\mathbb{R}^n$ \mathbb{R}^3$ (for $n=1$ you get a cylinder; for $n=2$ some sort of grid)grid; for $n=3$ it is sometimes called a jungle gym). $S^2$ and $S^3$ are the surfaces described by Richard Kent in his third paragraph. All these These two surfaces , for $n>1$, have exactly one end which is orientable but has genus. Therefore they all are homeomorphic. This is a pretty incredible result in my opinion. The most simple presentation of this surface is called the Loch-Ness monster: it is constructed by adding to a plane a sequence of handles placed in a row.
show/hide this revision's text 2 corrected a typo

As noted in the comment of Ryan Budney to Richard Kent answer, the non-compact surfaces have been classified (in a very nice way, hence my need to share the details).

I have no access to Jstor right now, so I rely on my memory and, well, you may want to check all this.

Ian Richards theorem says that non-compact surfaces (without boundary) are classified by there their orientablility, their genus (possibly infinite) and a triple of spaces, each one embedded in the preceding, that are:

  1. the space of its ends,

  2. the space of its ends with genus,

  3. the space of its unorientable ends.

The space of ends is constructed by taking an increasing sequence of compact subset that cover a topological space $T$, and looking at the connected components of their complements. An end of $T$ is an infinite, decreasing sequence of such connected components. The point is that you can do that in a way that does not substancially depend on the sequence of compact you chose.

For example, $\mathbb{R}^n$ has only one end provided $n\ge2$ (look at the sequence of balls of integer radius and centered at some point), while $\mathbb{R}$ has two ends. The space of ends of a regular tree is a Cantor set.

An end is said to have genus if the connected components that define it all have genus (they never reduce to annuli). An end is said to be unorientable if the connected components that define it all are unorientable.

Now, consider the surface $S^n$ defined as the boundary of a tubular neighborhood of the usual embedding of the usual Cayley graph of $\mathbb{Z}^n$ into $\mathbb{R}^n$ (for $n=1$ you get a cylinder; for $n=2$ some sort of grid). $S^2$ and $S^3$ are the surfaces described by Richard Kent in his third paragraph. All these surfaces, for $n>1$, have exactly one end which is orientable but has genus. Therefore they all are homeomorphic. This is a pretty incredible result in my opinion. The most simple presentation of this surface is called the Lock-Ness Loch-Ness monster: it is constructed by adding to a plane a sequence of handles placed in a row.
show/hide this revision's text 1

As noted in the comment of Ryan Budney to Richard Kent answer, the non-compact surfaces have been classified (in a very nice way, hence my need to share the details).

I have no access to Jstor right now, so I rely on my memory and, well, you may want to check all this.

Ian Richards theorem says that non-compact surfaces (without boundary) are classified by there orientablility, their genus (possibly infinite) and a triple of spaces, each one embedded in the preceding, that are:

  1. the space of its ends,

  2. the space of its ends with genus,

  3. the space of its unorientable ends.

The space of ends is constructed by taking an increasing sequence of compact subset that cover a topological space $T$, and looking at the connected components of their complements. An end of $T$ is an infinite, decreasing sequence of such connected components. The point is that you can do that in a way that does not substancially depend on the sequence of compact you chose.

For example, $\mathbb{R}^n$ has only one end provided $n\ge2$ (look at the sequence of balls of integer radius and centered at some point), while $\mathbb{R}$ has two ends. The space of ends of a regular tree is a Cantor set.

An end is said to have genus if the connected components that define it all have genus (they never reduce to annuli). An end is said to be unorientable if the connected components that define it all are unorientable.

Now, consider the surface $S^n$ defined as the boundary of a tubular neighborhood of the usual embedding of the usual Cayley graph of $\mathbb{Z}^n$ into $\mathbb{R}^n$ (for $n=1$ you get a cylinder; for $n=2$ some sort of grid). $S^2$ and $S^3$ are the surfaces described by Richard Kent in his third paragraph. All these surfaces, for $n>1$, have exactly one end which is orientable but has genus. Therefore they all are homeomorphic. This is a pretty incredible result in my opinion. The most simple presentation of this surface is called the Lock-Ness monster: it is constructed by adding to a plane a sequence of handles placed in a row.