Let f(x) be continuous on R and suppose f(x)< or equal to 1 for all x
rational.  Show that f(x) < or equal to 1 for all x in R

F(x) is defined as returning values less than or equal to 1 for all
rational numbers in the domain, D. Every rational number returns a
value less than or equal to 1, however, in between any two rational
numbers, another number can be found in between. By the definition of
the domain, this domain value would also return a value less than or
equal to 1.  Now, when we take a number in between, say, 9/3 and 10/3,
such as 3.2, we know that f(x) will return a value less than or equal
to 1. We can keep doing this infinitely and pick a number between 3
and 3.2, say, 3.14. In doing this, we see that any number selected in
between two numbers will result in a number less than or equal to 1.
3.1415 (~pi), can be represented by 31415/10000. Or more accurately,
31415926/10000000. Either of these numbers satisfies the original
domain, so the fact that we have to pick more and more accurate
numbers to approach the irrational, transcendental value of pi doesn’t
mean that irrational values won‘t return the values that are
approached by the rational values of the domain.  This is also proven
by the fact that the function is continuous, meaning that any value in
between the endpoints of the domain will return a value in the range,
whether rational or irrational. If the function is graphed, one can
see that if dots are to be drawn through two rational points and the
function is continuous there, an irrational point would have to return
a value less than or equal to 1 because two rational points that are
less than or equal to 1 surround the one irrational point.

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questions about it.

Thanks,
chis-ga

Let me fill in some details of the "nut and bolts" of this proof.

Suppose x is any real number.  We want to show that f(x) <= 1, given
that f is continuous on the real numbers R and that the inequality is
satisfied for any x in the rational numbers Q.

If x were rational, we would already have the desired conclusion.  But
even if x is irrational, we can still express x as the limit of a
sequence of rational numbers, e.g. the truncated decimal expansions of
x converge to it.  [Because every real number is the limit of a
sequence of rational numbers, we say that Q is a dense subset of R (in
the usual topology).]

Let {q_i} be a sequence of rational numbers converging to x:

x = LIMIT q_i [AS i --> +oo]

By virtue of the continuity of f, we then have:

f(x) = LIMIT f(q_i) [AS i --> +oo]

and each f(q_i) <= 1.  It follows by a standard argument that f(x) <=
1, which for the sake of completeness we will reproduce here.

Let any epsilon > 0 be given.  Then for all sufficiently large i, f(x)
is within epsilon of f(q_i), and in particular for some i depending on
epsilon we have:

|f(x) - f(q_i)| < epsilon

Therefore:

f(x) = f(q_i) + f(x) - f(q_i)

    <=   1  + |f(x) - f(q_i)|

     <    1  +  epsilon

But epsilon > 0 was arbitrary, so we have f(x) <= 1 "as epsilon tends
to zero".  More explicitly, if f(x) were greater than 1, picking:

epsilon = f(x) - 1 

would give the contradiction:

f(x) <  1 + epsilon  = 1 + f(x) - 1 = f(x)

Since f(x) is not greater than 1, then f(x) <= 1 as we were asked to
prove.

regards, mathtalk-ga