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While it is true that the product and addition/subtraction of continuous functions is continuous, is it true that the product and addtion/subtraction of discontinuous functions is also discontinuous?

Edit Thanks to Chris Eagle I now understand the example f-f, but I wonder under what conditions my above question/statement holds true?

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Consider $f-f$... – Chris Eagle Nov 14 '11 at 2:49
@ChrisEagle Thank you. Under what conditions does my above statement hold though? – analysisj Nov 14 '11 at 2:50
Arturo covered well all cases. The only thing to add is that different type of discontinuities cannot cancel eachother: if $f$ has a jump/removable/infinite or essential discontinuity at $a$ and $g$ has any other type of discontinuity at $a$ then $f+g$ is discontinuous at $a$. – N. S. Nov 14 '11 at 3:43
If you're familiar with linear algebra, continuous functions form a vector space $V_1$ over the reals. That vector space is a subspace of the vector space $V_2$ of all real functions. You're asking if $V_2 - V_1$ has any characteristics of a vector space, and in general the answer is no. – Dimitrije Kostic Nov 14 '11 at 4:40
up vote 6 down vote accepted

Let us say that a function is "discontinuous at $a$" if it is defined but not continuous at $a$.

Let $f$ and $g$ be real valued functions of real variable, and let $a$ be a real number. Here are a few things that are true:

  • If $f$ and $g$ are continuous at $a$, then so are $\alpha f$ (for any real number $\alpha$), $f+g$, $f-g$, $fg$, and if $g(a)\neq 0$ also $\frac{f}{g}$.

  • If $f$ is continuous at $a$ and $g$ is discontinuous at $a$, then $f+g$, $f-g$, and $\alpha g$ (for any real number $\alpha\neq 0$) are discontinuous at $a$.

  • If $f$ is continuous at $a$, $g$ is discontinuous at $a$, and $f(a)\neq 0$, then $fg$ is discontinuous at $a$.

  • If $g$ is discontinuous at $a$, and there is an open interval containing $a$ where $g$ is never equal to $0$, then $\frac{1}{g}$ is discontinuous at $a$.

For example, to show that if $f$ is continuous at $a$ and $g$ is discontinuous at $a$, then so is $f+g$, note that if $f+g$ were continuous at $a$, then $(f+g)-f$ would also be continuous at $a$. But $f+g-f = g$, which is discontinuous at $a$.

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