Let $S = \{f:\lim_{x\to 3}f(x)=0\}$. This set is a subspace of the vector space $\{f:f\colon\mathbb{R}\to\mathbb{R}\}$, $S$ being closed under addition and scalar multiplication. But what is the multiplicative identity in $S$? The constant function $1$ cannot belong to $S$ as its limit value will be one and not zero. Then will $S$ be a vector space? I'm not able to figure out where is that I'm being wrong. It'll be really helpful if someone can help me.
2026-04-01 03:05:23.1775012723
Multiplicative identity in a vectorspace
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Multiplication of vectors, such as via an inner product, or "cross" product in some cases, are all additional structures on a vector space. As the basic definition stands, one requires plenty of additive properties to hold : the vector space must be a group under addition.
On top of that, there is scalar multiplication, which is an operation that is supposed to behave well with addition. Therefore, vector spaces are not set up to even accommodate multiplication in general, let alone a multiplicative identity etc. All such structure will be obtained in algebras, which will combine well with the additive properties to give a strong structure to the set. In your case, $\{f : \lim_{x\to 3} f(x) = 0\}$ is a subspace simply because it is closed under addition (sum of limits is limit of sum), scalar multiplication (constant can be taken outside the limit) and the fact that zero is in the set.