### Continuity decoded

The definition of continuity was framed after decades of deliberation and mathematical squabbling. The current notation we have is due to a Polish mathematician by the name of Weierstrass. It states that

“If $f:\Bbb{R}\to \Bbb{R}$ is continuous at point $a$, then for every $\epsilon>0$, $\exists\delta>0$ such that for $|x-a|<\delta$, $|f(x)-f(a)|<\epsilon$.”

Now let us try and interpret the statement and break it down into simpler statements, in order to give us a strong visual feel.

Can $\epsilon$ be very large? Of course! It can be $1,000,000$ for example. Does there exist a $\delta$ such that $|x-a|<\delta\implies |f(x)-f(a)|<1,000,000$, even if the function is not continuous? Yes. An example would be $f(x)=x$ for $x\in(-\infty,a)$ and $f(x)=x+1$ for $x\in[a,\infty)$

Does this mean that we have proved a discontinuous function to be continuous? NO. $\epsilon$ should take up the values of all positive real numbers. So $f(x)$ defined above will fail for $\epsilon$ lower than $0.000\dots01$

Let us suppose for some $\epsilon>0$, we have $|f(x)-f(a)|<\epsilon$ if $|x-a|<\delta$. Let $f(x_1)$ and $f(x_2)$ be two points in $B(f(a),\epsilon)$. Let us now make $\epsilon=\frac{|f(x_1)-f(x_2)|}{2}$. Will the value of $\delta$ also have to decrease? Can it in fact increase?

The value of $\delta$ cannot increase because the bigger interval will contain $x_1$ and $x_2$, and we know that that will violate the condition that for all points in $B(a,\delta)$, the distance between the mappings has to be less than $\frac{|f(x_1)-f(x_2)|}{2}$. Can $\delta$ remain the same? No (for the same reasons, as the interval will still contain $x_1$ and $x_2$). Hence, $\delta$ most definitely has to decrease in this case?

However, does it always have to decrease? No. An example in case is a constant function like $y=b$.

We have now come to the most important aspect of continuity. The smaller we make $\epsilon$, the smaller the value of $\delta$. Does continuity also imply that the smaller we make $\delta$, the smaller the value of $\epsilon$? YES! How? When we decrease $\delta$, $\epsilon$ obviously can’t get bigger. Moreover, we know that there do exist values of $\delta$ which make smaller $\epsilon$ possible. Say, for $|f(x)-f(a)|<\epsilon/2$, it is necessary that $|x-a|<\delta/5$. Hence, if we decrease the radius of the interval on the x-axis from $\delta$ to $\delta/5$, the value of $\epsilon$ (or the bound of the mappings of the points) also decreases to $\epsilon/2$.

In summation, a continuous function is such that

decrease in value of $\epsilon\Longleftrightarrow$ decrease in value of $\delta$

One may ask how does knowing this help?

It has become very easy to prove that differentiable functions are continuous, and a host of other properties of continuous functions.

A doubt that one may face here is does this imply that all continuous functions are differentiable? No. “decrease in value of $\epsilon\Longleftrightarrow$ decrease in value of $\delta$” just implies that the derivative formula at $a$ will have a limit for every cauchy sequence of $x$ converging to $a$. In order for a function to be derivable, all those limits of the different cauchy sequences have to be equal. This is not implied by the aforementioned condition.