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Let $f$ be the function defined by $$f \colon \mathbb{R} \longrightarrow \mathbb{R}^{\omega}$$ $$f(t) = (t, 2t, 3t, ...)$$ Furthermore, let the topology on $\mathbb{R}$ and $\mathbb{R}^{\omega}$ be the square topology, i.e (in the case of $\mathbb{R}^{\omega}$) the topology induced by the metric $$\overline{p}(x, y) = \sup \{\overline{d}(x_{\alpha}, y_{\alpha})\}$$ Where $\overline{d}(x, y)$ is the standard bounded metric. Suppose I want to show this function is continuous.

I know it suffices to consider $f$ as follows: $$f(t) = (f_1(t), f_2(t), f_3(t) ...)$$ and show that $f_i(t)$ is continuous for all $i$. This is quite easy to do using an "$\epsilon$-$\delta$ argument". This style of proof has a marked analytic flavour to it, however, and I was wondering if there is a more topological way to approach a proof such as this.

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I don't understand what is the topology on $\mathbb R^\omega$. What is $J$ and what happens if the sequence of distance is unbounded in $\mathbb R$? – Asaf Karagila Feb 7 '13 at 17:37
@AsafKaragila Apologies, the $J$ was some notation I added out of habit. $R^{\omega}$ is the infinite Cartesian product $\mathbb{R} \times \mathbb{R} \times \mathbb{R} \dots$, i.e. the space of all infinite sequences. The topology is the one induced by the "square topology" which is the same as the one induced by the given (and now amended metric). – providence Feb 7 '13 at 22:19
I know what is $\Bbb R^\omega$. Now tell me, what is the distance of the sequences $x=(0,0,0,\ldots)$ and $y=(0,1,2,3,4,\ldots)$? – Asaf Karagila Feb 7 '13 at 22:21
@AsafKaragila OH! My goodness, I will change the definition of the metric given. I have made a mistake. – providence Feb 7 '13 at 22:22
Okay, that's better. Despite it not changing a single bit, does the "standard bounded metric" mean cut off at $1$ (i.e. if $|x-y|\geq 1$ then $\bar d(x,y)=1$); or is it $\frac{|x-y|}{1+|x-y|}$? – Asaf Karagila Feb 7 '13 at 22:29
up vote 2 down vote accepted

Let $(X_i)_{i \in I}$ be a family of topological spaces, $X= \prod\limits_{i \in I} X_i$ be the associated product space, $\pi_i : X \to X_i$ be the canonical projections ($i \in I$). Then, you can show that $f : Y \to X$ is continuous iff $\pi_i \circ f=f_i$ is continuous for every $i \in I$.

It is a direct consequence of the product topology, since you have the following commutative diagram:

$$\begin{array}{ccc} X_i & \leftarrow^{\pi_i} & X \\ \uparrow{f_i} & \nearrow{f} & \\ Y & & \end{array}$$

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The last paragraph of the question, starting "I know it suffices ...", would indicate that you use the product topology, in which case Seirios's answer applies. But earlier in the question, you referred instead to a certain "metric" $d$, which depends on an unspecified set $J$. If $J\neq\omega$ then your $d$ isn't a metric, because if $x_\alpha=y_\alpha$ for $\alpha\in J$ but not for $\alpha\in\omega-J$, then $d(x,y)=0$ even though $x\neq y$. On the other hand if $J=\omega$, then the topology induced by the metric $d$ is not the product topology and your function $f$ is not continuous from $\mathbb R$ to the metric space $\mathbb R^\omega$. The closest connection I can see between your "metric" formula and the product topology is that the product topology is induced by the infinite family of metrics $d$ that you get by taking all finite subsets of $\omega$ as $J$.

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I have updated the question. There were a few mistake, my apologies. – providence Feb 7 '13 at 22:34

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