Is there a locally compact, locally connected, Hausdorff and second countable space that is "nowhere locally Euclidean"?

Question

When I study topological manifold, I think some property of manifolds are so important that they can "almost characterize" manifolds. But I know a topological manifold is not easily to be characterized because of its locally Euclidean property.

I want to find a "weird" example of a locally compact, locally connected, Hausdorff and second countable space $X$ that is "nowhere locally Euclidean", i.e. for any point in $X$, there doesn't exist an open neighborhood that is homeomorphic to a open subset of a Euclidean space.

Since a discrete space is a 0-dimensional manifold(locally $\mathbb R^0$), the following type of example is not what I want:

A straight line union a point which is not on that line

Does such an example exist?

Every solution or reference or even more generalized example(Even stronger conditions can't guarantee the locally Euclidean property) will be appreciated!

Answer 1

Let $X$ be the Hilbert cube, which is $[0,1]^{\mathbb{N}}$ with its product topology. It is compact Hausdorff (by the Tychonoff theorem). Being a countable product of second-countable spaces, it is second countable. (In fact, it is metrizable, and every compact metric space is second countable). Moreover, it is a product of connected and locally connected spaces, so it is also connected and locally connected. (Note: a product of locally connected spaces is not in general locally connected, but it is when all the factors are also connected.)

To see $X$ is nowhere locally Euclidean, we use the invariance of domain theorem. Suppose to the contrary there exist nonempty open subsets $U \subset X$ and $V \subset \mathbb{R}^{\mathbb{n}}$ and a homeomorphism $f : U \to V$. Since $U$ is open, by definition of the product topology, there exist $x_1, \dots, x_m$ such that $\{(x_1, \dots, x_m)\} \times [0,1]^{\mathbb{N}} \subset U$. Let $W = (0,1)^n$ be the open unit cube in $\mathbb{R}^n$ and define a map $g : W \to U \subset X$ by $$g(y_1, \dots, y_n) = (x_1, \dots, x_m, y_1, \dots, y_n, 0, 0, \dots).$$

Clearly $g$ is continuous and injective. Moreover the image $g(W)$ is not open in $X$ nor in $U$. Since $f$ is a homeomorphism, $f(g(W))$ is not open in $V$ nor in $\mathbb{R}^n$. Yet $f \circ g : W \to V \subset \mathbb{R}^n$ is a continuous injection, so this contradicts invariance of domain.

Answer 2

The Menger sponge is an example. It is a 1-dimensional space into which every compact, metrizable, second countable, 1-dimensional space may be embedded.

In fact there similarly exist universal Menger compacta of every dimension, as was proved by Bestvina in his thesis, and these are all examples of what you ask for.

The Sierpinski carpet mentioned in the comments is another example, and it has an analogous universality property, with respect to compact, 1-dimensional planar spaces.

Added: To address the comment of @NateEldredge, the idea behind the proof of local connectivity is to look at the inductive intersection construction of the object itself --- say the Menger sponge, or the Sierpinski carpet --- and to see how this construction exhibits a neighborhood basis of each point, where each neighborhood basis element is itself a nested intersection of connected open sets in the ambient space, and can therefore be shown to be connected.

In the case of the Sierpinski carpet, one could look at Whyburn's paper "Topological characterization of the Sierpiński curve" (which I have not done), which proves that the Sierpinski carpet is characterized up to homeomorphism as the unique locally connected 1-dimensional continuum in the plane satisfying various additional properties. The Menger sponge and the higher dimensional Menger compacta have similar universal characterizations, and local connectivity is part of the characterization. In all cases, I am pretty sure that the proof of local connectivity will follow a similar scheme, using the inductive construction of the objects themselves.