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Given the topological space $X=\mathbb{R}^{[0,1]}$ with the product topology, there are several properties regarding to $X$ which I am not sure if are true/false.

  1. Is $X$ metrizable? I'm having trouble on how I can prove/disprove this, and I'm not sure if I should aim to proving this or to find a counterexample.

  2. Is $X$ normal and/or Hausdorff? I think I can show $X$ is Hausdorff, but not quite sure as to this one or on how to formalize such a proof.

  3. Is $X$ compact and/or locally compact? (And does compactness implies locally compactness or is it the other way around?)

  4. Is $X$ connected and/or path-connected?

I find it especially hard on how to start on proving/disproving each, so even a hint will help!:) Thanks!

Stefan Hamcke
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Zhan I.s.
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  • $X$ is not first countable (in particular not metrizable), $X$ is Hausdorff and its not normal. – azarel Oct 19 '14 at 14:16
  • I formatted your questions as a numbered list, and removed the [tag:algebraic-topology] tag which is not applicable. – Nate Eldredge Oct 19 '14 at 14:23
  • Every compact space $Y$ is locally compact. To show $Y$ is locally compact, we need to show that every point of $Y$ has a neighborhood which is compact. But if $Y$ is compact, for each point we can take that neighborhood to be $Y$ itself. Not every locally compact space is compact; $\mathbb{Z}$ is a counterexample. – Nate Eldredge Oct 19 '14 at 14:24

2 Answers2

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$X$ is not metrizable since it is not first-countable.

Since $\Bbb R$ is Hausdorff, so is an arbitrary product of $\Bbb R$. More generally, an arbitrary product of Hausdorff spaces is Hausdorff.

If $X$ were compact, another space $Y$ which you could write as the continuous image of $X$ would have to be compact too. Can you guess which map $X\to Y$ I'm talking about?
Local compactness is disproved similarly. Can you think of a reason why no neighborhood in $X$ can be compact?

$X$ is path-connected. Given points $x=(x_i)_{i\in[0,1]}$ and $y=(y_i)_i$ in $X$, a path can be expressed using the individual paths $x_i\to y_i$ over all $i\in[0,1]$.
Connectedness is a bit more difficult. The usual proof I know of uses the dense connected subset $D=\{(x_i)_i\mid x_i=p_i\text{ for almost all $i$}\}$ where $p=(p_i)_i$ is some fixed point in $X$.

Edit: Najib reminds me that path-connectedness implies connectedness, so the last paragraph applies if you wanted to show connectedness directly, or for the general case of a product of connected spaces.
In the same spirit, if you have disproved local compactness, then you've disproved compactness, since a compact Hausdorff space is always locally compact.

Stefan Hamcke
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The space $\mathbb{R}^{[0,1]}$ is just a product of $|\mathbb{R}|$ many copies of $\mathbb{R}$.

A product space $\prod_i X_i$ is connected iff all $X_i$ are connected. $\mathbb{R}$ is connected so.. The same theorem holds for path-connectedness as well (and is easier to prove even).

A product space $\prod_i X_i$ is compact iff all $X_i$ are compact. So the only question that remains: is $\mathbb{R}$ compact?

A product space $\prod_i X_i$ is Hausdorff iff all $X_i$ are Hausdorff. So is $\mathbb{R}$ Hausdorff?

Try showing that $\mathbb{R}^{[0,1]}$ is not first countable. This should kill illusions of metrisability.

The hardest to see is that $\mathbb{R}^{[0,1]}$ is not normal. For this see this question and its answers.

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Henno Brandsma
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