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Definitions: A function f with domain a subset D of \(\displaystyle R^{k}\) and range contained in \(\displaystyle R^{\ell}\), \(\displaystyle k = 1\) or \(\displaystyle 2\) and \(\displaystyle \ell = 1\) or \(\displaystyle 2\), is continuous at a point \(\displaystyle a \in D\) if for each \(\displaystyle \epsilon > 0\) there exists \(\displaystyle \delta > 0\) such that \(\displaystyle \forall x \in B_{\delta}(a) \cap D\), \(\displaystyle f(x) \in B \epsilon (f(a))\). The function is continuous on D if it is continuous at each point of D.

Let \(\displaystyle \epsilon > 0\) choose a \(\displaystyle \delta = \dfrac{\epsilon + 5}{8}\).

Let \(\displaystyle a \in R^{k}\) where \(\displaystyle k = 1\) or \(\displaystyle 2\). We can define an open ball \(\displaystyle B_{\delta}(a) = \){\(\displaystyle y \in R^{k}: d(y, a) < \delta\)}. Let \(\displaystyle x \in B_{\delta}(a) \cap R^{k}\).

Let \(\displaystyle b \in f(a)\). We can define another open ball \(\displaystyle B_{\epsilon}(b) = \){\(\displaystyle z \in R^{k}: d(z, b) < \epsilon\)}.

We need to prove that \(\displaystyle f(x) \in B\epsilon(b)\) which is equivalent to the statement \(\displaystyle B\epsilon(b) \cap f(x) = f(x)\). Since we want \(\displaystyle f(x)\) to be fully contained in the interval \(\displaystyle (f(a) - \epsilon, f(a) + \epsilon)\). This is equivalent to the statement \(\displaystyle f(x) < \epsilon\). \(\displaystyle f(x) = 8x - 5 < 8\delta - 5 = 8(\dfrac{\epsilon + 5}{8}) - 5 = \epsilon + 5 - 5 = \epsilon\).

Thus we have proven that the function \(\displaystyle f(x) = 8x - 5\) is continuous in \(\displaystyle R^{k}\).

Additional Questions:

We know that x is in the range of \(\displaystyle (a - \delta, a + \delta)\). So x can be more than delta if not centered at 0. So a slight mistake in the proof?

It really should be that \(\displaystyle f(x) < \epsilon\) is \(\displaystyle -\epsilon< f(x) < \epsilon\), so the proof is incorrect?

The interval \(\displaystyle (f(a) - \epsilon, f(a) + \epsilon)\) is only defined in \(\displaystyle R^{1}\). How do I define this concept in \(\displaystyle R^{2}\).

Let \(\displaystyle \epsilon > 0\) choose a \(\displaystyle \delta = \dfrac{\epsilon + 5}{8}\).

Let \(\displaystyle a \in R^{k}\) where \(\displaystyle k = 1\) or \(\displaystyle 2\). We can define an open ball \(\displaystyle B_{\delta}(a) = \){\(\displaystyle y \in R^{k}: d(y, a) < \delta\)}. Let \(\displaystyle x \in B_{\delta}(a) \cap R^{k}\).

Let \(\displaystyle b \in f(a)\). We can define another open ball \(\displaystyle B_{\epsilon}(b) = \){\(\displaystyle z \in R^{k}: d(z, b) < \epsilon\)}.

We need to prove that \(\displaystyle f(x) \in B\epsilon(b)\) which is equivalent to the statement \(\displaystyle B\epsilon(b) \cap f(x) = f(x)\). Since we want \(\displaystyle f(x)\) to be fully contained in the interval \(\displaystyle (f(a) - \epsilon, f(a) + \epsilon)\). This is equivalent to the statement \(\displaystyle f(x) < \epsilon\). \(\displaystyle f(x) = 8x - 5 < 8\delta - 5 = 8(\dfrac{\epsilon + 5}{8}) - 5 = \epsilon + 5 - 5 = \epsilon\).

Thus we have proven that the function \(\displaystyle f(x) = 8x - 5\) is continuous in \(\displaystyle R^{k}\).

Additional Questions:

We know that x is in the range of \(\displaystyle (a - \delta, a + \delta)\). So x can be more than delta if not centered at 0. So a slight mistake in the proof?

It really should be that \(\displaystyle f(x) < \epsilon\) is \(\displaystyle -\epsilon< f(x) < \epsilon\), so the proof is incorrect?

The interval \(\displaystyle (f(a) - \epsilon, f(a) + \epsilon)\) is only defined in \(\displaystyle R^{1}\). How do I define this concept in \(\displaystyle R^{2}\).

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