Let $f,g:\mathbb R \rightarrow \mathbb R$ be both continuous. Supose that $f(x)=g(x)$ for all $ x \in D$, where $D\subseteq\mathbb R$ is dense. Show that $f(x)=g(x)$, for all $x \in \mathbb R$. Id like a hint to solve this question.
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5Consider $h^{-1}({0})$ where $h(x) = f(x)-g(x)$. – Daniel Fischer Jul 13 '13 at 20:10
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Letting $\mathcal D$ be the set of dense subsets of $\Bbb R$, is the hypothesys $$(\forall D\in \mathcal D)(\forall x\in D)(f(x)=g(x))$$ or is it $$(\exists D\in \mathcal D)(\forall x\in D)(f(x)=g(x))?$$ – Git Gud Jul 13 '13 at 20:10
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Seems to me that it is the latter. – Nick Peterson Jul 13 '13 at 20:12
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A more general version of this question can be found here. – Cameron Buie Jul 13 '13 at 21:31
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Hint: For any $x_n \to x$ you have $f(x) = \lim_n f(x_n)$ and $\lim_n g(x_n) = g(x)$. For each $x$, is it possible to select a sequence $x_n$ such that $x_n \to x$ and $\lim_n f(x_n) = \lim_n g(x_n)$?
nullUser
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One more hint:
Define $h(x) = f(x)-g(x)$. For any given $y$ and $\delta$, there exists $x \in D$, such that $|x-y| < \delta$. Now, for prove that for any given $\epsilon$, since $h$ is continuous, $h(y) - h(x) = h(y) <\epsilon$.
Nathaniel Bubis
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Let $X$ be a topological space, and $Y$ a Hausdorff space. Then the set $\mathrm{diffker}(f,g)$ of points $x\in X$ such that $f(x)=g(x)$ is closed in $X$. For its complement is open: if $x\notin\mathrm{diffker}(f,g)$ then $f(x)\neq g(x)$ and $f(x)$ and $g(x)$ have distinct open neighborhoods in $Y$, taking the intersection of their preimages gives a neighborhood of $x$ in $X$ contained in $\mathrm{diffker}(f,g)$. If $\mathrm{diffker}(f,g)$ is dense in $X$, then $\mathrm{diffker}(f,g)=\overline{\mathrm{diffker}(f,g)}=X$. Now put $X=Y=\mathbf{R}$.
