I have a few questions about topological spaces which I am currently studying. First some definitions that I am using:

Definition of subspace topology: Given a topological space $(X,\tau)$ and a subset $S$ of $X$, the subspace topology on $S$ is defined by $\tau_{S} = \{S \cap U : U \in \tau \}$.

Definition of linear subspace: I use the usual definition, contains the zero vector and closed under addition and scalar multiplication.

Definition of topological vector space: A vector space $X$ over a field $K$ which is endowed with a topology such that vector addition $X \times X \rightarrow X$ and scalar multiplication $K \times X \rightarrow X$ are continuous functions(where $X \times X$ endowed with product topology).


  1. Why is the topological vector space defined in this way?
  2. Is a linear subspace in a topological vector space(e.g. normed space) automatically a subspace of the topological vector space or does it require additional properties?
  3. Does a subspace of a topological vector space satisfy the definition of subspace as stated in my first definition? How would I show this?
  4. Consider a normed vector space $X$ and a subset $W$. If we endow $W$ with a different norm that that of $X$ then what are the requirements necessary for $W$ to be a subspace of $X$?

Thanks a lot for any assistance!


You're using the word 'subspace' to mean two different things. When we talk about the 'subspace topology', we just mean endowing any subset of the topological space $X$ with a topology making it a topological space (hence, subspace). However, the definition of linear subspace is "subspace closed under the operations of linear algebra" - that is, a subspace of a vector space. Now, a topological vector space is two things at once - but above all, we always want it to be a vector space. So when we say the subspace of a topological vector space, we mean it in both ways at once - it's a subspace (in the sense of linear algebra) endowed with the subspace topology (making it into a topological space) - so a subspace of a topological vector space is also a topological vector space.

Re: #4: When we talk about a subspace of a topological vector space, we are specifically endowing a linear subspace with the subspace topology - so if the new norm induces a different topology, it's just another topological vector space, no mention made of our original one.

  • $\begingroup$ Thanks for your response. A couple of questions: So do you get vector subspaces of topological vector spaces which are not topological vector subspaces? Do you have an example with normed spaces? Why is the topological vector space defined in the way that the function $X \times X \rightarrow X$ and $X \times K \rightarrow X$ are continuous with respect to the product topology? $\endgroup$ – user103184 Dec 9 '13 at 14:19
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    $\begingroup$ Not quite. The difference between the two definitions is that a vector subspace has to satisfy certain properties to be a vector subspace - closed under addition and scalar multiplication. But a subspace of a topological space isn't something that satisfies properties - we just picked a subset and gave it a topology. Once you pick a vector subspace, it "automatically" gets a topology from the larger topological space. $\endgroup$ – user98602 Dec 9 '13 at 14:21
  • $\begingroup$ I see yes. Do you maybe know why the topological vector spaces are defined in this way? and it is specified in the definition that $K$ is a topological field, withou getting to much into the theory of fields, can this simply be though of as the real line with the usual topology which has base $\{ (a,b): b > a \text{ for }a,b \in \mathbb{R}\})$? $\endgroup$ – user103184 Dec 9 '13 at 14:37
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    $\begingroup$ We want a topological vector space to be defined this way because we want its operations (scalar multiplication and addition) to be continuous - this gives us access to all the tools of topology AND all the tools of linear algebra! $\endgroup$ – user98602 Dec 9 '13 at 14:40
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    $\begingroup$ Sure: consider the finite complement topology (a set is closed iff it has finitely many elements). I'll leave it to you to show that that's not continuous (find an open set whose inverse image isn't open). $\endgroup$ – user98602 Dec 10 '13 at 14:47

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