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I am having trouble understanding a proof in Rudin's "Real and Complex Analysis."

The theorem states that

To every linear transformation $T$ of $\mathbb R^k$ into $\mathbb R^k$ corresponds a real number $\Delta (T)$ such that $$m(T(E))=\Delta (T)m(E)$$ for every $E\in\mathfrak M$. In particular, $m(T(E))=m(E)$ when $T$ is a rotation.

Here $m$ is a complete measure defined on a $\sigma$-algebra $\mathfrak M$ such that $m(W)=vol (W)$ for every k-cell $W$.

(Part of) The Proof

Let $T:\mathbb R^k\rightarrow\mathbb R^k$ be linear...elementary linear algebra tells us that $T$ is a one-to-one map of $\mathbb R^k$ into $\mathbb R^k$ whose inverse is also linear. Thus $T$ is a homeomorphism of $\mathbb R^k$ into $\mathbb R^k$...

What I don't understand

I am not quite convinced that the linearity of $T$ is enough to prove that $T$ is a homeomorphism. From what I found, a linear mapping is continuous if the space is a normed finite dimensional space. However, here we are not guaranteed that $\mathbb R^k$ is finite dimensional. So wouldn't the proof be incomplete? What am I missing? What allows $T$ be a homeomorphism?

hjhjhj57
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CoffeeIsLife
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    What makes you think that $\mathbb{R}^k$ is not finite-dimensional? – Damian Reding Jun 07 '15 at 01:30
  • Because it is not assumed so. – CoffeeIsLife Jun 07 '15 at 01:31
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    A couple of things: First, the proposition is non-trivial only when $T$ is an isomorphism, why? Second, $\mathbb{R}^k$ is the canonical example of a real $k$-dimensional vector space. – hjhjhj57 Jun 07 '15 at 01:31
  • In all of maths there is nothing more finite-dimensional than $\mathbb{R}^k$, and the dimension is $k$. – Damian Reding Jun 07 '15 at 01:33
  • Yes, I understand that it is non trivial when it is an isomorphism since the volume would be zero otherwise. However, why can we assume that $\mathbb R^k$ is finite? Can't $k=\infty$? – CoffeeIsLife Jun 07 '15 at 01:34
  • You need to go back to your favorite linear algebra textbook. You don't assume $\mathbb{R}^k$ is finite dimensional. You know it has a group structure under adition and that you can define a product by real scalars such that the axioms of a vector space are satisfied. Then, it is a real vector space. Can you give a basis for it? I don't think the book is considering the case for $k=\infty$. – hjhjhj57 Jun 07 '15 at 01:36
  • I am not assuming that it is. Are you both misreading homeomorphism with homomorphism...? – CoffeeIsLife Jun 07 '15 at 01:42
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    every linear operator between two finite dimensional vector spaces is continuous: http://math.stackexchange.com/questions/112985/every-linear-mapping-on-a-finite-dimensional-space-is-continuous – hjhjhj57 Jun 07 '15 at 01:45
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    It goes without saying that the notation $\mathbb{R}^k$ implies that $k$ is a natural number (and $\infty$ is not a number) – Damian Reding Jun 07 '15 at 01:54

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First observe that $\mathbb{R}^k$ is a finite dimensional real vector space, a basis for it being the set $\{e_i\}_{i=1}^k$, where all the components of $e_i$ are zero except the $i$th one, which is $1$. (For example, if $k=3$ the set $\{(1,0,0),(0,1,0),(0,0,1)\}$ is a basis for $\mathbb{R}^3$.)

Now, since $T:\mathbb{R}^k\to\mathbb{R}^k$ is an isomorphism between two finite dimensional vector spaces and every linear function between finite dimensional vector spaces is continuous, you know that $T$ is an homeomorphism with inverse $T^{-1}$.

Edit: Let me know if something isn't clear.

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