Hi everyone,(adsbygoogle = window.adsbygoogle || []).push({});

I came across something in my vector calculus textbook (Marsden and Tromba, Edition 5, p. 327) that is confusing me.

"A function f(x,y) is said to beboundedif there is a number M>0 such that -M<=f(x,y)<=M for all (x,y) in the domain of f. A continuous function on a closed rectangle is always bounded, but, for example, f(x,y) = 1/x on (0,1]x[0,1] is continuous but is not bounded, because 1/x becomes arbitrarily large for x near 0. The rectangle (0,1]x[0,1] is not closed, because the endpoint 0 is missing in the first factor."

I do not see how f(x,y) = 1/x could possibly be considered bounded on the closed rectangle [0,1]x[0,1] by the definition given - it still approaches infiniti for x near 0. There is clearly no M for which f(x,y) <= M for all (x,y) in the domain of f, be it on [0,1]x[0,1] or (0,1]x[0,1]. And yet, it is stated, without explanation, that the function is bounded on this rectangle.

I looked at some other sources to try to find a different definition of bounded that might clear up some of the confusion, but to no avail (Wikipedia has an equivalent definition and also says that a continuous function on a closed region is always bound).

Does anyone know what the deal is?

Thanks,

HJ Farnsworth

**Physics Forums - The Fusion of Science and Community**

# Simple but frustrating confusion on definition of bounded function

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Simple but frustrating confusion on definition of bounded function

Loading...

**Physics Forums - The Fusion of Science and Community**