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What are the components and path components of $\mathbb{R}^\omega$ (in the product topology)?

At the moment, I am just working on the first part of the question. I read somewhere online that a space $X$ has only one component if and only if $X$ is connected. Since $\mathbb{R}^\omega$ is connected, then according to this fact $\mathbb{R}^\omega$ is the only component. I suspect the same is true of path-connectedness, but hopefully someone will kindly correct me if this is wrong.

Unfortunately, this fact about connectedness hasn't been presented in the book I am working through. How could one solve this without that fact; I must confess, I have really no intuition about components yet, especially in the infinite dimensional space $R^ω$. I could use some hints.

As a way of knowing what ideas/theorems I have available to me, this is problem 2a in section 25 of chapter 3 in Munkres' Topology.

EDIT:

Theorem 25.1: The components of $X$ are connected disjoint subspaces of $X$ whose union is $X$, such that each nonempty connected subspace of $X$ intersects only one of them.

I believe this is a proof: If $X$ only has one component, and it is suppose to partition $X$, then every element of $X$ must be in this single component, proving that $X$ equals this single component; moreover, since components are connected, it follows that $X$ is connected.

Now, suppose that $X$ is connected but has at least two components. Since components are nonempty, each has at least one element. But $X$ is a connected subspace containing both of these elements, and it certainly intersects both components, which is at odds with theorem 25.1. Hence, $X$ can only have one component.

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    $\begingroup$ Look at the definition of "component". How can you have two such sets when $X$ is connected? And if $X$ is disconnected, there must be non-empty disjoint open sets $A, B$ with $A\cup B = X$. if $a \in A, b\in B$, then the component containing $a$ and that containing $b$ cannot be the same, so there must be at least two components. $\endgroup$ – Paul Sinclair May 3 '17 at 0:55
  • $\begingroup$ @PaulSinclair Hmm. What you are suggesting isn't obvious to me. My book gives the following definition about components: "Given $X$, define an equivalence relation on $X$ by setting $x \tilde y$ if there is a connected subspace of $X$ containing both $x$ and $y$.The equivalence classes are called components of $X$" In addition to this, I also know that components are themselves connected. It seems that the only way it is impossible for a connected space $X$ to have at least two components is if these components are also open--that seems to be the only way to obtain a contradiction. $\endgroup$ – user193319 May 4 '17 at 18:08
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    $\begingroup$ If $X$ is connected, then $X$ itself is a connected subspace of $X$ that contains $x$ and $y$. $\endgroup$ – Paul Sinclair May 4 '17 at 20:40
  • $\begingroup$ @PaulSinclair Okay. I think have a proof based up on your remark; I made an edit on my original post. $\endgroup$ – user193319 May 5 '17 at 19:29
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    $\begingroup$ That works, though it is more simple to say: If $X$ is connected, then for any $x, y \in X, X$ itself is a connected subspace containing both $x, y$. Therefore $x \sim y$. Hence the component of any $x \in X$ is all of $X$. QED. More generally, note that if $A$ is any connected subset of an arbitrary space $X$, then any two elements of $A$ share a component (because $A$ provides the needed subspace for their equivalence). So a connected set always lies in a single component (though that component may also contain other points). $\endgroup$ – Paul Sinclair May 5 '17 at 19:44

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