To ensure
Definition: ''A smooth manifold is a locally ringed space $(M;C^{\infty})$ which satisfies the Hausdorff condition, you can say conditions:
Paracompactness can be stated by saying that the
Paracompactness and second countability are not equivalent conditions on locally euclidean Hausdorff spaces (second countability implies paracompactness, but not vice versa, see
Explanations:
1) is evident.
And
2) means that for $x \neq y \in M$, there exists $f \in C^{\infty}(M)$ with $f(x)=0$, $f(y) \neq 0$. This ensures the Hausdorff condition once we know that elements of $C^{\infty}(M)$ give rise to continuous maps $M \to \mathbb{R}$. This is as follows: $f \in C^{\infty}(M)$ given, $x \in M$has only countably many connected components . Pick a chart $h:U \to \mathbb{R}^n$; under this chart, $f|_U$ corresponds to a smooth function on $\mathbb{R}^n$, whose value at $h(x)$ does not depend on the choice of the chart. Call this value $f(x)$. Checking the continuity of $x \mapsto f(x)$ can be reformulated by saying done in charts.
3.) By this I mean that for each open cover $C^{\infty}(M)$ has only countably many indecoposable idempotents (U_i)$, there is a partition of unity $\lambda_i$ with the usual properties and that $\lambda_i$ is a map of $C^{\infty} (an M)$-modules. A standard argument shows that $\lambda_i$ is given by multiplication with a smooth function. Therefore, the underlying space $M$ is paracompact.
4.) An idempotent $p\neq 0 $ in a commutative ring $A$ is called indecomposable if
holds. Indecomposable idempotents in $C^{\infty}(M)$ correspond to connected components.
So here is the definition:
Definition: ''A smooth manifold is a locally ringed space $(M;C^{\infty})$ which satisfies the conditions:
These conditions together imply that $(U,C^{\infty})$ M$ is isomorphic to $(\mathbb{R}^n,C^{\infty})$ as Hausdorff and second countable, because a locally ringed euclidean, connected and paracompact Hausdorff space .

