Show that the Euclidean topology in a product of spaces Vector matches the product of Euclidean topologies. A topological isomorphism between two topological vector spaces is an application between the two that is both isomorphism and homeomorphism. If $V$ is a vector $\mathbb K$-space of finite dimension $n$ over a metric field $K$, we will call the Euclidean topology in $V$ to the only topology with respect to which all isomorphisms of $V$ in $\mathbb {K}^n$ are topological.
A metric field is a pair $(K, | |)$, where $K$ is a field and $| |$ is an absolute value in K.
An absolute value in K is any function $||:K\rightarrow R$ , where R is an Archimedean ordered body, which fulfills the following properties:
a) $| x | ≥ 0$ and the only element of $K$ that satisfies $| x | = 0$ is $x = 0$.
b) $| x + y | ≤ | x | + |y|$
c) $| xy | = | x || y |$
The advice to prove it is to follow this: "If $W$ is a subspace of $V$, then the restriction to $W$ of the topology Euclidean in $V$ is the Euclidean topology."
How can I show that?
This is what I have tried:
The definition of euclidean topology is defined in finite vectors space, so we will assume $\prod_{i\in I}V_i$ finite. So our product of vector space will be denoted by $\displaystyle\prod_{i=1}^{i=n}V_i$
Considering:
*$\mathcal T_1$ is the euclidean topology of $\displaystyle \prod_{i=1}^{i=n}V_i$
*$\mathcal T_2=T_{a_1}\times ...\times T_{a_t}$ is the product topology of $\displaystyle\prod_{i=1}^{i=n}V_i$ where $T_{a_i}$ is the euclidean topology of $V_i$.
*Each $V_i$ has dimension $n_i$.
Then we have to show that $f:(\displaystyle\prod_{i=1}^{i=n}V_i,\mathcal T_1)\longrightarrow (\prod_{i=1}^{i=n}V_i,\mathcal T_2)$ is an homeomorphism.
$t:(\displaystyle\prod_{i=1}^{i=n}V_i,\mathcal T_1)\longrightarrow \mathbb {K}^t$ when $t=\displaystyle\sum_{i=1}^{i=n}n_i$ is an homeomorphism (by defition of euclidean topology).
Also $g:\mathbb {K}^t\longrightarrow \mathbb {K}^{n_1}\times...\times\mathbb {K}^{n_n}$ is an homeomorphism. If we show that
$h:(\displaystyle\prod_{i=1}^{i=n}V_i,\mathcal T_2)\longrightarrow {K}^{n_1}\times...\times\mathbb {K}^{n_n}$ is an homeomorphism then we will have demonstrated it.
Since each $(V_i,\mathcal T_{a_i})$ is homeomorphic to $\mathbb K^{n_i}$. Then each projection $\pi_i:{K}^{n_1}\times...\times\mathbb {K}^{n_n}\longrightarrow (V_i,\mathcal T_{a_i})$ is continouos. Also each $\pi'_i=h\circ \pi_i:(\displaystyle\prod_{i=1}^{i=n}V_i,\mathcal T_2)\longrightarrow (V_i,\mathcal T_{a_i})$ is continous because is the projection. Then $h$ is continouos.
Since each $(V_i,\mathcal T_{a_i})$ is homeomorphic to $\mathbb K^{n_i}$ and considering each $\pi'_i$ definded above we can defined $\pi''_i:(\displaystyle\prod_{i=1}^{i=n}V_i,\mathcal T_2)\longrightarrow \mathbb K^{n_i}$ that will be continous. Then each $h^{-1}\circ \pi''_i:{K}^{n_1}\times...\times\mathbb {K}^{n_n}\longrightarrow \mathbb K^{n_i}$ is continouos because is projection. Then $h^{-1}$ is continous. Then $h$ is an homeomorphism (I have not mentioned it previously, but it is bijective because each $V_i$ is homeomorphic to $K^{n_i}$)
. Then $f$ is an homeomorphism too.
 A: Let me give you another hint as to how to break the problem down:
Let us show inductively that for $$\mathbb{R}^n = \mathbb{R} \times \cdots \times \mathbb{R}$$ the euclidean (=norm topology) on the left hand side coincides with the the product topology on the right hand side. (Note the more classical approach is to show that the euclidean norm is equivalent to the maximum norm, however I will strictly use topological arguments working with open sets)
Argue by induction over dimension $n\geq 2$. Starting with $n = 2$. We have $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$ as a set. Due to my comment it suffices to prove that

*

*every $0$-neighborhood in $\mathbb{R}^2$ contains a neighborhood of the form $U \times V$, where $U,V$ are $0$-neighborhoods in $\mathbb{R}$

*every neighborhood of the form $U \times V$, where $U,V$ are $0$-neighborhoods in $\mathbb{R}$ contains a norm ball $B_\varepsilon^{\mathbb{R}^2}(0) = \{(x,y) \in \mathbb{R}^2 \mid x^2+y^2 < \varepsilon^2 \}$. (superscript denotes the euclidean space I take the balls in)

For 1. If $W$ is a euclidean $0$-neighborhood in $\mathbb{R}^2$ we may assume without loss of generality that $W = B_\varepsilon^{\mathbb{R}^2}(0)$ (why?)
Then we choose $\delta>0$ so small that $2\delta^2 < \varepsilon^2$ and thus
$B_\delta^{\mathbb{R}}(0) \times B_\delta^{\mathbb{R}}(0) \subseteq B_\varepsilon^{\mathbb{R}^2}(0)$ and this proves 1.
For 2.: If $U\times V$ is a $0$-neighborhood in the product topology we may shrink $U,V$ so that we may assume without loss of generality that $U=B_\delta^{\mathbb{R}}(0)=V$ for some $\delta>0$, but then $B_\delta^{\mathbb{R}^2}(0) \subseteq U \times V$ (write this down explicitely!).
So this proves 2. and we have seen that the product topology coincides with the euclidean topology on $\mathbb{R}^2$.
Assume now that we have proven the claim for all $2\geq n$ and we wish to continue with $n+1$.
Observe first that this implies that we can always replace $\mathbb{R}^m = \mathbb{R}^{m-k} \times \mathbb{R}^k$ (since we know that the product topology coincides with the euclidean one on all lower dimensional spaces! Again here I am hiding details which one should work out).
Then we write $\mathbb{R}^{n+1}= \mathbb{R}^n \times \mathbb{R}$ and it suffices to prove that the norm topology on the left coincides with the product topology from the right (by the observation we just made). Now you can almost verbatim copy the argument from the induction start to see that the topologies coincide.
I hope this helps, now you only need to transport the topologies using your isomorphisms (and this is in principle only notation).
Again: A lot of this problem gets easier by just directly proving that the euclidean norm is equivalent to the maximum norm (and indeed to every other norm on the finite dimensional subspace). So norms are your friends, but I understood that you were looking for a purely topological argument.
