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Let $Y=\oplus_{i \in N}Y_i$ and $Y_i$ is the metriable space $[0,1]$ for each $i \in N$. $x_n \in Y$ denotes that $x_n \in Y_n$ and $0\le x_n \le 1$. $d$ is the usual metric on $[0,1]$. We define a function $f$ from $Y^2$ to $[0,1]$ as following:

$f(x_n,y_m)=d(x_n,y_m)$, if $n=m$; otherwise, $f(x_n,y_m)=\frac{1}{n}+\frac{1}{m}$.

Is this function $f$ continuous? And how to prove it?

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Which metric you defined on $Y$ and $Y^2$ –  no identity Dec 22 '11 at 7:13
    
I guess that you are considering the usual topology on $\mathbb{R}$. But for $Y^{2}$ there are a lot of topology we could consider. So what topology in $Y^2$ you want to consider ? –  Leandro Dec 22 '11 at 7:17
    
The topology on $Y$ is generated by the sum of $Y_i$. The topology on $Y^2$ is the Tychonoff topology, i.e., the topology is generated by the Cartesian product of the space $Y$. –  Paul Dec 22 '11 at 8:11

2 Answers 2

up vote 1 down vote accepted

Since Y is metrizable, you can also prove it via convergent sequences.

For notational convenience I’ll write $Y=\omega\times[0,1]$. A sequence $\left\langle\langle n_k,x_k\rangle:n\in\omega \right\rangle$ in $Y$ converges to $\langle n,x\rangle$ iff $\langle x_k:k\in\omega\rangle\to x$ in $[0,1]$ and $n_k=n$ for all sufficiently large $k$, so a sequence $$\Big\langle\big\langle\langle n_k,x_k\rangle,\langle m_k,y_k\rangle\big\rangle:k\in\omega\Big\rangle$$ converges to $\big\langle\langle n,x\rangle,\langle m,y\rangle\big\rangle$ in $Y^2$ iff $\langle x_k:k\in\omega\rangle\to x$ in $[0,1]$, $\langle y_k:k\in\omega\rangle\to y$ in $[0,1]$, and $n_k=n$ and $m_k=m$ for all sufficiently large $k$. If $n\ne m$, the sequence $$\Big\langle f\big(\langle n_k,x_k\rangle,\langle m_k,y_k\rangle\big):k\in\omega\Big\rangle$$ is eventually constant with value $\frac1n+\frac1m$, and if $n=m$ it’s eventually just the sequence of $|x_k-y_k|$, which certainly converges to $|x-y|$. Thus, $f$ preserves convergent sequences and is therefore continuous.

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The connected components of $Y^2$ are of the form $Y_i \times Y_j$. All of these components are (closed and) open, so they form an open cover of $Y^2$. Since the restriction of $f$ to each component in the cover is continuous, $f$ itself is also continuous.

To prove this from first principles, consider what the preimage $U$ of an open subset $V$ of $[0,1]$ under $f$ must look like. (Specifically, consider the intersection of $U$ with each of the connected components mentioned above. You'll find that they're all open, meaning that $U$ is a union of open sets, and thus open.)

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I'd have done it from first principles,but the connected component method works since by definition, the union of the connected components is the whole space in the range. Well done. –  Mathemagician1234 Dec 22 '11 at 8:51
    
Indeed, the important thing is actually that, since $Y^2$ is locally connected, the connected components form an open cover. On a space with non-open connected components a different approach would be needed. –  Ilmari Karonen Dec 22 '11 at 9:29
    
@Iimari VERY good point indeed! It drives home the point that it's absolutely critical to know exactly which sets in a given topological space are open when considering the continuity of functions.You'd be surprised how many students don't do this carefully when beginning to work with general notions of continuity outside of Euclidean spaces. –  Mathemagician1234 Dec 22 '11 at 9:35
    
@ Ilmari Karonen,do you mean that the function $f$ is continuous? It seems a little complex for me. –  Paul Dec 22 '11 at 9:56
    
@John: Yes, unless I've made some silly mistake, $f$ is continuous. You may want to tweak the definition of $f$ so that its values actually lie in $[0,1]$, though (or extend its codomain to all of $\mathbb R$, or at least to $[0, 1.5]$). –  Ilmari Karonen Dec 22 '11 at 11:00

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