Continuity of a Function on Disconnected Domains

Continuity of real functions: Survey of methods

There are two ways to check continuity - global and local.

Global continuity.

We know that great many functions are continuous on their domains, namely all elementary functions. We also know that when combining elementary functions by operations, continuity is preserved. Thus we can tell continuity of any function that is composed of elementary functions.

Example: The function

is continuous on its domain D(f )=(−∞,5) ∪ (5,∞), since on this domain it is created by combining continuous elementary functions.

Note that for continuity we need both-sided neighborhoods, so the above procedure gives continuity only on open sets. If the domain happens to include some of its endpoints, we only get one-sided continuity there. This may have far reaching consequences in more complicated setups (like split functions), so it is safer to use this global approach only on open sets, that is, to handle endpoints individually (see local continuity below).

Split functions. Split functions are defined by different formulas on (typically) disjoint sets. If every formula is continuous on the set to which it applies, then the split function is continuous on the union of interiors of these sets. Indeed, it is important to recall that when a function is continuous on an interval (a,b] and on an interval (b,c), then it need not be continuous on their union (a,c)! The reason is simple, continuity on (a,b] only means continuity at b from the left, we have no information about continuity from the right at b from this interval, and we also have no information about this right continuity at b from the second interval, since b does not belong to (b,c). These "connecting points" have to be handled locally, see the next part.

Local continuity.

Given a function f and a point a in its domain, we want to know whether f is continuous at a. There are two possibilities.

Case 1. There is some neighborhood U of a on which the function is defined by some expression whose continuity we already know. Then f is also automatically continuous at a. For instance, the function f from the previous example is continuous at a = 13.

Case 2. There is no such neighborhood. In the most typical case this means that the function is defined in different ways to the right and to the left of a. For instance, the absolute value |x| considered at a = 0 has this property, since it is defined by x to the right and by -x to the left from 0.

In such a case we determine continuity and classify discontinuity if it appears in the following way:

Step 1. Determine both one-sided limits at a and the number f (a).

Step 2. Compare the three numbers from Step 1 and draw conclusions. There are these possibilities:
• The two one-sided limits converge, they are equal (so they also give the (both-sided) limit) and they are also equal to f (a). Then the function is continuous at a
• The two one-sided limits converge and they are equal (so they also give the (both-sided) limit), but not equal to f (a). Then the function is not continuous at a, it has a removable discontinuity there.
• The two one-sided limits converge, but they are not equal. Then the (both-sided) limit does not exist, the function is not continuous at a, it has a jump discontinuity there.
• At least one of the one-sided limits does not converge. Then the function is not continuous at a, it has a discontinuity of the 3rd kind there.

One-sided continuity

To decide whether a given function is continuous at a from the left, we first evaluate its limit at a from the left and f (a), then we compare them.

If the limit from the left converges and is equal to f (a), then the function is continuous at a from the left. Otherwise the function is not continuous at a from the left. We do not classify discontinuities for one-sided continuity.

Continuity from the right is done in analogous way.

Example: Determine continuity of

Solution: We start with global continuity the easy way. For every interval specified in the definition, the function f is defined by one concrete continuous function on that interval, so it is also continuous there. Since we have to be careful and only work with open sets, we have the conclusion that the given function is continuous on the set

(−∞,−π) ∪ (−π,0) ∪ (0,1) ∪ (1,2) ∪ (2,∞).

What about the points missing from this set? We have to use the local approach and check on continuity at the points -π, 0, 1, and 2. To this end we find one-sided limits and function values at these points. Note that when doing one-sided limits, we have to use corresponding formulas.

Conclusions:
• At a = −π there is a jump discontinuity, but f is continuous from the left there.
• At a = 0 f is continuous.
• At a = 1 f is continuous.
• At a = 2 there is a discontinuity of the 3rd kind, but f is continuous from the right there.

When we put this together with the global step, we get the

Result: f is continuous on the set

(−∞,−π) ∪ (−π,2) ∪ (2,∞).

It is also continuous from the left at -π and continuous from the right at 2.

The picture shows what is happening, but note that we did not need it to make our conclusions.

For more examples see Solved problems.

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1-1 and inverse
Back to Methods Survey - Basic properties of real functions

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Source: https://math.fel.cvut.cz/mt/txtb/3/txe3bb3e.htm

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