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How does one show this?

Show that the set of all continuous functions $f:[0,1] \rightarrow R$ is a closed subset of set of all functions $g:[0,1] \rightarrow R$, when endowed with the box topology.

What does it mean if a set is endowed with a box topology? What function will fit this? What happens if I change the box topology to product topology?

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One way to think about the box topology on the space of all functions $f:[0,1]\to\mathbb{R}$ is that the basic open neighborhoods of such an $f$ consist of all functions that lay inside a tiny tube around $f$, whose radius can vary (even discontinuously).

Now, the point is that if every such tube around $f$ contains a continuous function, then $f$ itself must be continuous, by a simple $\epsilon/3$ argument. That is, for any $x\in[0,1]$ and any $\epsilon\gt 0$, pick a tube of radius $\epsilon/3$ and a continuous function $g$ within that tube. Since $g$ is continuous at $x$, there is some $\delta\gt 0$ such that any $y$ within $\delta$ of $x$ has $g(y)$ within $\epsilon/3$ of $g(x)$, and also $f(y)$ within $\epsilon/3$ of $g(y)$ and $f(x)$ within $\epsilon/3$ of $g(x)$. So altogether, $f(y)$ is within $\epsilon$ of $f(x)$, as desired.

In the product topology, in contrast, the basic open sets can be thought of as specifying tiny windows at finitely many points in the domain, such that any function that goes through those windows is counted as in the open set. (In the box topology, in contrast, we in effect get to specify a window at every point in the domain.) Consider the functions $f_n(x)=x^n$. The pointwise limit of these functions is not continuous, since it has value $0$ on $[0,1)$ and value $1$ at $x=1$. But any basic open neighborhood of that limit function will contain some $f_n$, and so the limit function is in the closure of the continuous functions, but not itself continuous.

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In other words, the box topology is finer than the uniform topology, and your argument shows that the continuous functions are already closed in the uniform topology. –  Chris Eagle Feb 9 '11 at 18:53
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In fact, the continuous functions are dense in the product topology, and this does not only hold for the unit interval, but for all Tychonov spaces. –  Henno Brandsma Feb 9 '11 at 19:01

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