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Suppose $f: \mathbb{R} \to \mathbb{R}$ is continuous function and consider the set $\{ (x,f(x)) : x \in \mathbb{R} \}$ = G. Then, we obviously know that this set $G$ is closed. Now, I am having some difficulty trying to find its interior. My guess is that $\operatorname{Int} G = \varnothing $. But, how can I prove this? Any help would be greatly appreciated.


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The interior of a set is open. Can you find an $p=(x,y)\in G$ such that $B(p,\epsilon)\subseteq G$ for some $\epsilon > 0$? –  David Peterson Feb 9 at 6:15
Let $(x, f(x))$ be an element in the graph and $B$ is a ball of any radius centered at that point. Then $(x, g(x)+ \epsilon)$.... –  John Feb 9 at 6:15
How about using a space-filling curve and extending it continuously all over the Real line? –  user99680 Feb 9 at 6:22
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3 Answers 3

up vote 1 down vote accepted

I think you need the function (EDIT: curve; please see below) to be injective; otherwise you have cases like that of a space-filling curve :http://en.wikipedia.org/wiki/Space-filling_curve . If you want to turn it into a function $f: \mathbb R \rightarrow \mathbb R$ , then you can extend continuously left and right. The existence of the continuous extension is guaranteed, e.g., by Tietze extension theorem.

EDIT: like was pointed out, this is not actually a function, but a curve (and I wrongly assumed you meant a curve, not a function). For a function, consider a ball about the pair $(x,f(x))$. Since f is a function, any ball $B((x,f(x));r)$ for any $r>0$ cannot contain any horizontal strip above $f(x)$, so the image must contain an empty interior. Say there is a closed ball $B$ in the graph. This ball will then contain a small vertical strip , which implies that some point in the domain has two images, which cannot happen for a function.

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I think you're confusing image with graph. –  nonpop Feb 9 at 6:38
If $f$ is a function its graph is quite far from a space-filling curve. –  David Peterson Feb 9 at 6:39
Yes, I wrongly assumed the OP meant a curve, not a function; I just edited. –  user99680 Feb 9 at 6:47
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The projection $\pi:G\to\Bbb R$ is a homeomorphism. Can you show $\Bbb R\times\{0\}\subset \Bbb R^2$ has empty interior?

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This set has measure $0$ in $\mathbb{R}^2$, so its interior must be empty.

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