Proof for Closure of Vector Addition - Can You Help?

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dEdt
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If the tangent space at p is a true vector space, then it must be that the sum of two vectors is itself a directional derivative operator along some path passing through p. I've been trying to prove that this is true without any luck.

My textbook "proves" that vector addition is closed by showing that when the sum of two vectors is applied to the product of two scalar functions, we just get the Leibniz rule. This argument seems really unconvincing to me.

Can anyone either justify the validity of my text's proof, or offer their own proof for the closure of vector addition?
 
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dEdt said:
If the tangent space at p is a true vector space, then it must be that the sum of two vectors is itself a directional derivative operator along some path passing through p. I've been trying to prove that this is true without any luck.

My textbook "proves" that vector addition is closed by showing that when the sum of two vectors is applied to the product of two scalar functions, we just get the Leibniz rule. This argument seems really unconvincing to me.

Can anyone either justify the validity of my text's proof, or offer their own proof for the closure of vector addition?

In the book by John Baez and Javier Muniain, "Gauge Fields, Knots and Gravity", the authors say that the easiest way to go about it is to start with a vector field, which assigns a vector to each point in spacetime. You can easily(?) prove that vector fields form a vector space, and then (maybe?) the fact for vectors follows.
 
What book are you learning from? The proof you're asking for will be in every textbook on differential topology. If you're learning differential topology from a physics book then my best advice would be to do yourself a favor and learn it from a proper book on the subject. See, for example, section II.5 of "Topology and Geometry"-Bredon (bear in mind this book defines smooth manifolds by using sheafs but it's equivalent to the usual definition in terms of smoothly compatible charts).