The following is a part from P.13 in Topological Vector Spaces (Third Printing Corrected 1971) by H.H.Schaefer

Given a vector space $L$ over a (not necessarily commutative) non-discrete valuated field $K$ and a topology ${\mathfrak T}$ on $L$, the pair $(L,{\mathfrak T})$ is called a topological vector space (abbreviated t.v.s.) over $K$ if these two axioms are satisfied :

$$(LT)_1 \;\; (x,y) \rightarrow x+y \;\; \mbox{is continuous on} \;\; L \times L \;\; \mbox{into} \;\; L $$ $$(LT)_2 \;\; (\lambda,x) \rightarrow \lambda x \;\; \mbox{is continuous on} \;\; K \times L \;\; \mbox{into} \;\; L $$

Then, 3 lines below follows the following sentence.

Since, in particular, this implies the continuity of $(x,y) \rightarrow x-y$, every t.v.s. is a commutative topological group.

Isn't any vector space commutative by definition ? Or, is it about the commutativity of $K$ ? But then, how is it shown from the continuity of the mapping ? I'm confused.

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    $\begingroup$ The continuity of $(x, y) \mapsto x - y$ implies that a t.v.s is a topological group; it says nothing about commutativity (which is automatic regardless of whether $K$ is commutative; remember that this refers to the multiplication of $K$, not the addition). $\endgroup$ – Qiaochu Yuan Oct 17 '11 at 15:23
  • $\begingroup$ Dear @QiaochuYuan. Thank you for your comment. My problem is solved. But still, it is not entirely clear to me. By $(LT)_1$ and $(LT)_2$, the group operations have already been proved to be continuous. I understand that the continuity of $(x,y)\rightarrow x-y$ is equivalent to $(LT)_1$ and $(LT)_2$. So, it seems redundant to me. $\endgroup$ – Aki Oct 17 '11 at 15:42
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    $\begingroup$ $(LT)_2$ is a stronger statement than the statement that negation is continuous; the author just wants to observe that it implies that negation is continuous. $\endgroup$ – Qiaochu Yuan Oct 17 '11 at 15:58
  • $\begingroup$ Thank you again @QiaochuYuan ! I believe I understand entirely. $\endgroup$ – Aki Oct 17 '11 at 16:29

The discussion of $x-y$ is redundant, as OP suspected. Without any continuity assumptions, vector addition and negation satisfy the axioms of a commutative group, and this is known from the basic theory of vector spaces not considering topology. By LT1 and LT2 these maps are continuous, so that vector addition in a topological vector space defines a commutative topological group.

The book author is referring to the idea that the structure of a group can be determined from its subtraction operation $S(x,y)=xy^{-1}$ and that the formulas for doing so show that if $S(x,y)$ is continuous, so are the addition and multiplication maps. Defining a group as a set with a subtraction operation is usually done as a short-cut for checking that some structure is a group, to avoid checking the multiplication and inverse axioms separately. In this case one already knows that there is a group structure and the only point to verify is continuity. Although the argument in the first paragraph of this answer is correct, giving the proof by subtraction is an exercise in using the axioms, handles the group law and continuity at the same time, and provides a one-line argument that might reassure those readers who would not be satisfied with a more logically economical (but longer to write down) observation that the proof follows from prerequisites.


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