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$\textbf{Textbook defenition:}$ Let $A$ be a subset of $\mathbb{R}$. Let $f: A \rightarrow \mathbb{R}$ and let $c \in A$. We say that $f$ is continuous at $c$ if, given any number $\epsilon>0$ there exists $\delta>0$ such that if $x$ is any point of $A$ satisfying $\mid x-c \mid<\delta$, then $|f(x)-f(c)|< \epsilon$.

In this definition, we have the freedom to choose $\epsilon$, that is, we sort of close in on $f(c)$ by the y-axis. Would the defenition still work if we have the freedom of choosing $\delta$? That is,

Let $A$ be a subset of $\mathbb{R}$. Let $f: A \rightarrow \mathbb{R}$ and let $c \in A$. We say that $f$ is continuous at $c$ if, given any number $\delta>0$ there exists $\epsilon>0$ such that if $x$ is any point of $A$ satisfying $\mid x-c \mid<\delta$, then $|f(x)-f(c)|< \epsilon$.

In this case, we are closing in on $f(c)$ from the $\text{y-axis}$. Would this work? If it doesn't then why?

MathFail
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Ellie_Wong
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  • Every bounded function is continuous according to your definition. – Kavi Rama Murthy May 30 '23 at 05:07
  • @geetha290krm How is that so? Can you elaborate – Ellie_Wong May 30 '23 at 05:10
  • Consider a unit step function. $f(x) = 0$ if $x\le0$ and $f(x) = 1$ if $x>0.$ This function is not continuous. Yet, for any $\delta> 0, |f(0+\delta) - f(x)| \le 1$ So, $\epsilon$ (slightly greater than $1$) exists, and does not force this function to be continuous. In the traditional definition, $\epsilon$ can be arbitrarily small. Your revised definition does put that same constraint on $\epsilon.$ – user317176 May 30 '23 at 05:47

2 Answers2

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Let $A$ be a subset of $\mathbb{R}$. Let $f: A \rightarrow \mathbb{R}$ and let $c \in A$. We say that $f$ is continuous at $c$ if, given any number $\delta>0$ there exists $\epsilon>0$ such that if $x$ is any point of $A$ satisfying $\mid x-c \mid<\delta$, then $|f(x)-f(c)|< \epsilon$.

Let $f(x)=0, x<0$, $f(x)=1, x\ge 0$, according to your definition,

For any $\delta>0$, there exists $\epsilon=2$, such that whenever $|x-0|<\delta$, $|f(x)-f(0)|<2$

it is "continuous" at $x=0$, but obviously this is wrong, since $f$ is not continuous at $x=0$.

MathFail
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Human translation for the first definition

$f$ is continuous at $c$ if and only if we can make $f(x)$ as close to $f(c)$ as we want, by choosing $x$ that is close enough to $c$.

If you want $f(x)$ to be extremely close to $f(c)$ then you need to choose $x$ that is also really close to $c$. From here the dependency is clear; you can choose an arbitrary $\epsilon$ and $\delta$ depends on $\epsilon$.

Human translation for the second clause

If we choose $x$ to be within certain distance from $c$, we guarantee that $f(x)$ will be within certain distance from $f(c)$.

acat3
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  • And what's faulty about the second clause? – Ellie_Wong May 30 '23 at 05:42
  • @Ellie_Wong the first definition guarantees that the difference $f(x)-f(c)$ can be extremely small, the second clause only guarantees that the difference will be finite, not necessarily small. Take for example $f(x)=1$ in $[-1,1]$ and $0$ everywhere else, this $f$ fits the second clause because $f(x)-f(c)<2$ for any $x$ and $c$ but it does not fit the first definition – acat3 May 30 '23 at 05:49