$\{0,1\}^{\mathbb{N}}$与$\mathbb{R}$是否同胚？含拓扑条件的技术问询

Let's walk through your two topology questions step by step—they're related, but the key difference in topological setup changes the answer entirely:

1. Are ${0,1}^{\mathbb{N}}$ and $\mathbb{R}$ homeomorphic? #

First off, let's clarify a critical point: homeomorphism is a property of topological spaces, not just plain sets. So the answer here depends entirely on what topologies we assign to each set:

• If we can choose any topologies: Since both sets have the same cardinality (they're both size $\mathfrak{c}$, the cardinality of the real numbers), we can absolutely make them homeomorphic. Just take a bijection between them, and transfer the topology from one space to the other. For example, if we give both the discrete topology, any bijection works as a homeomorphism. Or map the standard Euclidean topology from $\mathbb{R}$ onto ${0,1}^{\mathbb{N}}$ via a bijection—boom, they're homeomorphic.
• If we're referring to their "natural" standard topologies: That's exactly the setup of your second question, so let's dive into that in detail below.

2. Given a bijection $\psi:{0,1}^{\mathbb{N}}\to \mathbb{R}$, with ${0,1}$ having the discrete topology and ${0,1}^{\mathbb{N}}$ the product topology—are they homeomorphic? #

Short answer: No, they are not homeomorphic, even though there's a bijection between them. The topological structures are fundamentally different, and we can prove this using topological invariants—properties that are preserved by homeomorphisms:

• Compactness: By Tychonoff's theorem, ${0,1}^{\mathbb{N}}$ with the product topology (since ${0,1}$ is compact in its discrete topology) is a compact space. But $\mathbb{R}$ with the standard Euclidean topology is definitely not compact—think about the open cover ${(-n, n) \mid n \in \mathbb{N}}$; you can't cover all of $\mathbb{R}$ with any finite subset of these intervals. Since compactness is preserved by homeomorphisms, the two spaces can't be homeomorphic.
• Connectedness: Another decisive invariant. $\mathbb{R}$ is connected (it's a single interval, and intervals are connected in the Euclidean topology). But ${0,1}^{\mathbb{N}}$ with the product topology is totally disconnected. Here's what that means: take any two distinct sequences $x = (x_n)$ and $y = (y_n)$—there's some index $k$ where $x_k \neq y_k$. The set of all sequences where the $k$-th element is $x_k$ is both open (it's a basic open set in the product topology) and closed (its complement is the set of sequences with $k$-th element $y_k$, which is also open). So we can separate any two points with clopen sets, making the space totally disconnected. A totally disconnected space can't be homeomorphic to a connected one—this rules out any homeomorphism between the two.

Even though the sets are the same size (via the bijection), homeomorphism requires the bijection to preserve all topological properties, which just isn't possible here.

内容的提问来源于stack exchange，提问作者aleio1