# Comparison of two final topologies

Consider the vector space $F$ of all infinite sequences of reals numbers, such that only finitely many terms of each sequence are nonzero.

I recently encountered an exercise where I was required to show that

-the final topology on $F$ with respect to the injection maps $i_k:\mathbb{R}\rightarrow F, x\mapsto (0,...,0,x,0,...)$, where $x$ is situated at the $k$'th position ($k\in\mathbb{N}$) and $\mathbb{R}$ carries the usual topology

is identical to

-the final topology with respect to the inclusion maps $h_k:E_k\hookrightarrow F$, ($k\in\mathbb{N}$), where $E_k$ is that linear subspace in $F$ that consists of all sequences such that only the first $k$ entries of each sequence are nonzero (all the remaining ones are zero) and can thus be identified with $\mathbb{R}^k$ and also carries the topology corresponding $\mathbb{R}^k$ .

This seems to me to be false. Let us denote the first topology (the on with the injections) with $\tau_1$ and the second with $\tau_2$. Then the set $G:=\{ (x,0,0,...):x\in (0,1)\}$ is open in $\tau_1$, since $i_1^{-1}(G)=(0,1)$ is open in $\mathbb{R}$ and $i_k^{-1}(G)=\emptyset$ for all $k\geq 2$ is also open in $\mathbb{R}$.
But $G$ isn't open in $\tau_2$, because $h_2^{-1}(G)=G$, but this set isn't open in $E_2$ because it corresponds to $(0,1)\times \{0\}$ in $\mathbb{R}^2$ and that set isn't open.

Another thing that puzzles me is whether this space $F$ (endowed with the topology $\tau_1$) is Frechet: The problem seems to be that $F$ does not admit a countable subbase of seminorm on it that generates $\tau_1$, but I don't know how to prove this (we defined Frechet space to be those vector spaces that have a topology that is generated by countable subbase of seminorm and that is Cauchy-complete). Is there maybe a different (elementary) way of showing that it isn't Frechet ?
The images of the $i_k$ do not even cover $F$, so the first topology will be discrete outside the union of the axes. In particular, the point $(1,1,0,0,0,\dots)$ is open in $\tau_1$. –  Stefan Hamcke Apr 9 '14 at 15:49
@StefanHamcke Why exactly does it follow from that fact that the images of the $i_k$ don't cover $F$, that every open set outside the union of the axes is open in $\tau_1$ ? Because the preimages of those sets are always the empty set ? (It might a priori still be the case, that $\tau_2$ is the same as $\tau_1$) –  pink_pyjamas Apr 9 '14 at 15:56
Yes, exactly. We usually want $\bigcup f_i(X_i)$ to cover $Y$ (where $f_i:X_i\to Y$). Otherwise any set avoiding that union will have only empty preimages, so it will be open and closed. Especially when there is only one map $f:X\to Y$, we call it a quotient map if $Y$ has the final topology and $f$ is surjective. –  Stefan Hamcke Apr 9 '14 at 16:02