Continuity decoded

by ayushkhaitan3437

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.

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