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a. For every real number $c$, neither $f$ nor $g$ is continuous at $c$.

b. For every real number $c$, the sum $f + g$ is continuous at $c$.

c. For every real number $c$, the product $fg$ is continuous at $c$.

Hint: consider functions defined one way on rationals and another on irrationals.

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    This looks like a homework question.2010-09-23
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    Shall I delete my answer? But then we should also close the question because ii seems impossible to me to give a hint that is more concrete than the hint in the question but doesn't give a full answer.2010-09-23
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    Admittedly, it's a test practice question. I was just wondering if the question wanted us to give a piecewise function as an example.2010-09-23
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    P.Burdell. I think it would be nice that, even asking homework questions / test practices, everyone doesn't make just a copy & paste that a robot can do. I mean: it's nice to see some words from a human being at the beginning of the question, like: "Hi, this is homework, but could you give me some hints, please?"2010-09-23
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    I just took the test. I think I made an A! Thank you stackexchange! @Agusti Sorry. I was getting anxious. It's my first test in college.2010-09-23

2 Answers 2

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Take $f=1_\mathbb Q$ (that is, $f(x)=1$ if $x$ is rational and $f(x)=0$ otherwise) and $g=1-f$. Then neither $f$ nor $g$ is continuous at any point but $f+g$ and $fg$ are very simple.

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This isn't specific to the real numbers. The domain, range, sum and product could be specified (almost) arbitrarily.

Let $S= f+g$ and $P=fg$. These are continuous functions. $f$ and $g$ are roots of a quadratic equation with continuous coefficients, $z^2 - Sz + P$. There is a formula expressing the solution continuously in terms of $S$ and $P$, using one extraction of a square root.

The only possible source of discontinuity is in the selection of which sign of the square root is used to define $f$ and which one for $g$. To attain discontinuity everywhere, one needs to have $S^2 \neq 4P$ everywhere (so that $f,g$ are always distinct) and for the set where the plus sign is used for determining $f$ to be dense with dense complement. Hence, a splitting of the domain into two dense sets (such as rational and irrational) is not an artificial feature of the answer posted earlier, it is necessary and sufficient for the solution.