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Problem Statement: Let $I = [a,b]$ and let $f: I \rightarrow \mathbb{R}$ and $g: I \rightarrow \mathbb{R}$ be continuous functions on $I$. Show that the set $E = \{x \in I: f(x) = g(x)\}$ has the property that if $(x_n) \subseteq E$ and $x_n \rightarrow x_0$, then $x_0 \in E$.

Here's what I have so far, I think it's complete, however I'd like to see a different proof if possible. Maybe one using the difference function, $lim[(f-g)(x_n)]$.

$Proof:$ Since $(x_n) \subseteq E$ and $f, g$ are continuous on $I$, then $limf(x_n) = f(x_0)$ and $limg(x_n)=g(x_0).$ Therefore $\forall \varepsilon >0.\ \exists \delta_1 >0$ such that $$|f(x_n) - f(x_0)| < \varepsilon /2$$ and similarly $\exists \delta_2 >0$ such that $$|g(x_n) - g(x_0)| < \varepsilon /2.$$ Choose $\delta = inf \{\delta_1, \delta_2 \}. \forall x \in E \cap V_\delta(x_0)$ \begin{align*} |f(x_0) - g(x_0)| & = |f(x_0) - f(x_n) + g(x_n) - g(x_0)| \ & \ [\text{Since}\ f(x) = g(x)\ \forall x \in E]\\ & \le |f(x_n) - f(x_0)| + |g(x_n) - g(x_0)| \\ & < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{align*} Since $\varepsilon$ is arbitrarily small $f(x_0) = g(x_0)$, therefore $x_0 \in E. \square$

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Let $h=f-g$. Then $h$ is continuous, so $\lim h(x_n)=h(x_0)$. –  Gerry Myerson Nov 13 '13 at 12:17

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Since each $x_n$ lies in the closed set $[a,b]$ and $x_n\to x_0,$ then $x_0\in [a,b],$ and so $h(x_0)$ is defined. (You left this out of your proof.)

Now, let $h=f-g.$ Note that $h$ is a continuous function such that $h(x_n)=0$ for all $n,$ so since $x_n\to x_0,$ then....

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