TITLE: "Proof" that $g(t) = (t,t,t,...)$ is not continuous with uniform topology
QUESTION [5 upvotes]: Let $g : \mathbb R \to \mathbb R^{\omega}$ be the function
$$
 g(t) := (t, t, t, \ldots).
$$
If $\mathbb R^{\omega}$ is equipped with the uniform topology, and $\mathbb R$ with the standard topology, then is $g$ continuous? According to this post
 it is. I have a different proof which yields another conclusion, but I cannot see whats wrong with the proof?
Let $Y = \prod_{i=1}^{\infty} Y_i \subseteq R^{\omega}$, then
$$
 g^{-1}(Y) = \bigcap_{i=1}^{\infty} Y_i.
$$
Proof: Let $x \in g^{-1}(Y)$, then $g(x) = (x,x,x,\ldots) \in Y$ and so $x \in Y_i$ for all $i$, thus $x \in \bigcap_{i=1}^{\infty} Y_i$. Conversely let $x \in \bigcap_{i=1}^{\infty} Y_i$, then $g(x) = (x,x,x,\ldots) \in \prod Y_i = Y$, and so $x \in g^{-1}(Y)$. $\square$
A basis set $B(x,r)$ in the uniform topoloy on $\mathbb R^{\omega}$ has the form
$$
 B(x,r) = \{ y = (y_i) : |x_i - y_i| < r \mbox{ for all }i \}
        = \prod_{i=1}^{\infty} B(x_i, r) = \prod_{i=1}^{\infty} (x_i - r, x_i + r)
$$
so in particular for $x = (1,1/2,1/3,\ldots)$ the set
$$
 B(x, 1) = \prod_{i=1}^{\infty} \left( \frac{1}{n} - 1, \frac{1}{n} + 1 \right)
$$
is open in the uniform topology. But
$$
 g^{-1}(B(x,1)) = \bigcap_{i=1}^{\infty} \left( \frac{1}{n} - 1, \frac{1}{n} + 1 \right) = (0,1]
$$
is not open, so $g$ is not continuous? What goes wrong here....

REPLY [5 votes]: $B(x,r)$ is not open in uniform topology. For example take $y = (1, 1/2, 1/3, \dots)$. Then $z := (1, 1, 1, \dots) \in B(y, 1)$ but clearly $B(z, \epsilon) \not \subset B(y, 1)$ for every $\epsilon > 0$.