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Let $(x_n)$ be a sequence in $\mathbb{R^2}$ and $c \in \mathbb{R^2}$.

To show $f$ is continuous we want to show if $(x_n) \to c$, $f(x) \to f(c)$.

As $(x_n) \to c$ we can take $B_\epsilon(c)$, $\epsilon > 0$ such that when $n \geq$ some $N$, $x_n \in B_\epsilon(c)$.

As $x_n \in B_\epsilon(c)$ this implies that $f(x_n) \in f(B_\epsilon(c))$.

This holds for all $\epsilon$, so as $\epsilon \to 0$ and $B_\epsilon(c)$ becomes infinitely small, we can always find $n \geq$ some $N$ such that $x_n \in B_\epsilon(c)$ and $f(x_n) \in f(B_\epsilon(c))$.

Hence as $\epsilon \to 0$, $(x_n)$ clearly converges to $c$ and $f(x_n)$ clearly converges to $f(c)$.

Does that look ok?

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up vote 4 down vote accepted

Well, since you don't use the definition of $f$ anywhere, you could substitute any function, and this would prove that it's continuous. What you need to show is that for any $\varepsilon>0$ there exists an $N$ such that $f(x_n)\in B_\varepsilon (f(c))$, instead of $f(x_n)\in f(B_\varepsilon (c))$, for every $n\ge N$.

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Why is do we need to show $f(x_n)\in B_\varepsilon (f(c))$, instead of $f(x_n)\in f(B_\varepsilon (c))$? – sonicboom Nov 6 '12 at 20:20
Because if $x_n\in B_\varepsilon(c)$ then $f(x_n)\in f(B_\varepsilon(c))$ always whereas the first version is only true for continuous functions. If $f$ is not continuous, then even for a small $\epsilon$, $f(B_\varepsilon(c))$ might include points far away from $f(c)$ but that's not allowed for continuous functions. – nonpop Nov 6 '12 at 20:27
I am showing that the sequence $(x_n)$ in $\mathbb{R^2}$ converges to $c$ as the open ball $B_\epsilon(c)$ becomes infinitely small due to $\epsilon \to 0$. The image of this ball, $f(B_\epsilon(c))$ in $\mathbb{R}$ will also become infinitely small as $\epsilon \to 0$ and as this ball contains $f(x_n)$, $f(x_n)$ will converge to $f(c)$. Surely $\epsilon \to 0$ forces the image open ball to become infinitely small? – sonicboom Nov 6 '12 at 20:43
The image of a ball doesn't have to be open (even for continuous functions) and it doesn't have to be a ball. For non-continuous functions it doesn't have to become small. Consider the function $f(x,y)=0$ if $x<0$ and $f(x,y)=1$ if $x\ge 0$. Then for every $\varepsilon>0$, the image of the ball $B_\varepsilon(0)$ is the set $\{0,1\}$. This is not open, it's not a ball, and the diameter is always $1$. – nonpop Nov 6 '12 at 20:49
Thanks for the example. I get that now. – sonicboom Nov 6 '12 at 20:54

Purely topological proof: This proof works for any space $X$. In your case $X=\mathbb{R}$. You want to prove that the first projection $f:X \times X \to X$ is continuous. A function $f$ is continous if the inverse image of any open is open. So let $U \subseteq$ X be any open. Then the inverse image $f^{-1}(U)=U \times X$, which is open by definition of the product topology (its base consists of $V \times W$ with $V,W$ open in $X$).

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There's a bit of repetition when you say $x_n \in B_\epsilon(c) \implies f(x_n) \in f(B_\epsilon(c))$. While this is true as you define it, repeating it doesn't add to the proof. What you need to show is that the image of $B_\epsilon(c)$ is itself an open neighborhood of $f(c)$.

Another look, which uses uniform continuity...

For any open neighborhood $N_\delta(\mathbf{x})$ of $\mathbf{x} = (x_1,y_1) \in \Bbb R^2$ and $(x_2,y_2) = y \in N_\delta(\mathbf{x})$ with the usual metric we have

$$\delta > d(\mathbf{x},\mathbf{y}) = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \ge \sqrt{(x_1-x_2)^2} = |x_1-x_2| = d(f(\mathbf{x}),f(\mathbf{y})).$$

Thus, we can set $\delta = \varepsilon$ and obtain uniform continuity on any open subset of $\Bbb R^2$.

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