Are $X=M \times [0,T]$ and $\partial X$ smooth compact manifolds when $M$ is smooth compact Riemannian manifold?

Question

Let $X=M \times [0,T]$, where $M$ is a smooth and closed compact Riemannian manifold.

I want to know if: $X$ is smooth compact manifold, and if $\partial X$ is smooth compact manifold?

1. I am not sure if $X$ is compact. I think it is compact iff it has finite diameter and is geodesically complete. But take two points $a =(m,0)$ and $b=(m,T)$ where $m \in M$. Then the geodesic cannot be extended forever, right? Or have I misunderstood??

Also I guess $X$ is smooth since $M$ is.

2. Well $\partial X = M \times \{0, T\}$ which again I think is smooth for the same reason. I don't know about compactness.

Answer 1

Compactness is a topological property and should not depend on the metric at all. $X$ is compact, because is topologically a product of two compact spaces. Similarly for $\partial X=M\times \{0,T\}$.

Answer 2

The product $X = M \times [0, T]$ is compact, but it's not a manifold. (It's a manifold-with-boundary, which is a special type of non-manifold.) That's the resolution of the apparent problem with extending geodesics. Even if $M$ is a point, $X \simeq [0, T]$ isn't geodesically complete; this doesn't contradict your claim in 1. because $[0, T]$ isn't a manifold.