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I have the following problem : Let $S =(0,1) \times (0,1) \subset \mathbb{R} ^{2} $ and for each $s,0\leq s \leq 1$ let $V_{s} = {s}\times (0,1)$ and $f_{s} : V_{s} \rightarrow \mathbb{R} ,(s,t) \rightarrow t.$Does this make S into a one-manifold? First I thought of giving a sequence that converges on $V_{s} $ for example is $s=0$ chose the sequence $(n^{-1} )$ To see that it does not converge under the $f_{s} $ function but now I realize that it is like an identity in $\mathbb{R} $ that makes me confused.

Kevin
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  • It depends on your definition of a manifold and on what topology on $R^2$ you are considering. – Moishe Kohan Feb 27 '20 at 12:57
  • I guess it's the one induced by the euclidian norm, so $S$ has the relative topology. – Kevin Feb 27 '20 at 13:02
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    Then you should decide if $V_s$ is an open subset of $S$. – Moishe Kohan Feb 27 '20 at 13:03
  • Do some research and it's an exercise in Marsden's manifolds tensor analysis and applications book where the following definition appears: Let $M$ be a differentiable manifold. $A$ subset $A \subset M$ is called open if for each $a \in A$ there is an admissible local chart $(U,f)$such that $ a \in U$ and $U \subset A$. – Kevin Feb 27 '20 at 13:51

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If your intent is for the collection $\{(f_s, V_s)\}$ to serve as an atlas on $S$, then you should be more explicit about what topology you are giving $S$. In particular, the sets $V_s$ are not even open in $S$ if it is endowed with the topology it inherits as a subspace of $\mathbb{R}^2$ (or, equivalently, the topology it inherits from the Euclidean metric). The sets $V_s$ are, however, open if you give $S$ the order topology. The problem here is that manifolds are often required to be second countable, but $S$ in the order topology is not second countable.

Mark
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  • Thanks a lot for his answer, with the definition that comes in the book where the exercise is I'm clear: Let $M$ be a differentiable manifold. $A$ subset $A \subset M$ is called open if for each $a \in A$ there is an admissible local chart $(U,f)$such that $ a \in U$ and $U \subset A$. – Kevin Feb 27 '20 at 13:56
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    I think you've got something backwards. That definition explains how to tell if a subset of a manifold is open or not, but your questions is asking whether or not $S$ is a manifold in the first place. Go back and find how your text defines manifolds. A common definition is that a differentiable manifold is a topological space that is Hausdorff, second countable, locally Euclidean, and has differentiable transition maps. – Mark Feb 27 '20 at 14:25