Show that inclusion function $i:\mathbb{R}^\omega\to\mathbb{R}^\infty$ is continuous but not an homemorphism in its image. We define for each $n\in\mathbb{N}$ the subspace $\tilde{\mathbb{R}_n}\subset\mathbb{R}^\omega$ the space of sequences (with product topology) by: $\overline{x}=(x_k\in \tilde{\mathbb{R}_n})$ if and only if $x_k=0$ for all $k>n$. Let $\mathbb{R}^\infty=\bigcup_\limits{n\in\mathbb{R}}\tilde{\mathbb{R}_n}$ with the topology given by: $U\subset\mathbb{R}^\infty$ is open if $U\cap\tilde{\mathbb{R}_n}$ is open for all $n\in\mathbb{N}$.

*

*Show that a topology is defined in $\mathbb{R}^\infty$

*Show that the inclusion function $i:\mathbb{R}^\infty\to\mathbb{R}^\omega$ is continuous but it isn't an homeomorphism in its image.

I was able to solve item 1, directly satisfying the topology definition. In item 2, it was trivial that inclusion function is continuous since it's an identity with an extended codomain and if we restrict the codomain to its image, we have $j:\mathbb{R}^\infty\to i(\mathbb{R}^\infty)=\mathbb{R}^\infty$ the new inclusion function. It's easy to see that is injective (since $i$ was), surjective since we restrict the domain to its image and continuous since $i$ was continuous. We also know that the inverse of this function, $j^{-1}$ ,exists.
Hence what fails in the homeomorphism definition is the continuity of $j^{-1}$.
Then one possibility is picking an open set $U\in\tau_\infty$ (the topology we got in item a) and see it's preimage $j^{-1}(U)$ is not open.\
I thought about using a constante sequence, $(1)_n$ which is not in $\mathbb{R}^\infty$ because there are no $0$'s $\forall n\in\mathbb{N}$ but since we have subspaces, relative topologies like $\tau_{p/\infty}=\{U\cap\mathbb{R}^\infty:U\in\tau_p\}$ ($\tau_p$ is the product topology we said for $\mathbb{R}^\omega$) i'm very confused and can't conclude why $j$ is not an homeomorphism.
 A: First point:  your proof that $i$ is continous is incorrect. The identity map is not always continuous, since the topology of the domain might be different from the topology of the co-domain.
I'll let you ponder a bit longer on the continuity of $i$ while I try to explain why its inverse (defined on the range) is not continuos.
Consider the set
$$
U=\mathbb R^\infty\cap \prod_{k\in\mathbb N}\big(-1,1\big).
$$
The intersection of $U$ with each $\tilde{\mathbb R}_n$ is the open set
$\prod_{k=1}^n(-1,1)$, so $U$ is open in $\mathbb R^\infty$.  However $i(U)$ is not open in the range of $i$ (can you prove it?), so $i$ is not an open map (onto its range) which means that its inverse is not continuous.

EDIT: Here is why $i(U)$ isn't open in $i(\mathbb R^\infty)$.  Supposing otherwise, we'd have that $\mathbb O := (0,0,0,\ldots)$ is an
interior point of $i(U)$.  Therefore there exists an open subset $\Omega $ of $\mathbb R^\omega $ such that
$$
  \mathbb O \in  \Omega \cap i(\mathbb R^\infty) \subseteq  i(U).
  $$
If so,  then  $\Omega $ may clearly be taken as a basic open set, namely a set of the form
$$
  \Omega = A_1\times A_2\times \dots\times A_n\times \mathbb R\times \mathbb R\times \mathbb R\times \dots
  $$
where the $A_i$ are open subsets of ${\mathbb R}$, for every $i=1, 2, \ldots , n$.  Since $\mathbb O \in  \Omega $,  we have that $0\in A_i$, for
all $i$, and hence
$$
  (0,0,\ldots ,\underbrace{0}_{\text{n.th position}}, 2, 0, 0, \ldots ) \in   \Omega \cap i(\mathbb R^\infty) \subseteq  i(U),
  $$
but this is clearly not true.
