Bolzanos Theorem

**mitwhiz** · May 18th, 2010, 02:33 PM

I want to use Bolzanos Theorem to prove that a function changes sign on an interval from -inf to inf. However, this theorem is defined for closed intervals, and the infinity interval is an open one. Is there an obvious and easy way to prove this for an open interval, or in the special case of (-inf, inf)?

**Drexel28** · May 18th, 2010, 04:29 PM

Originally Posted by mitwhiz

I want to use Bolzanos Theorem to prove that a function changes sign on an interval from -inf to inf. However, this theorem is defined for closed intervals, and the infinity interval is an open one. Is there an obvious and easy way to prove this for an open interval, or in the special case of (-inf, inf)?

I assume what you mean is that $\lim_{x\to-\infty}f(x)\leqslant y\leqslant \lim_{x\to\infty}f(x)$ .

**mitwhiz** · May 18th, 2010, 06:26 PM

Let me rephrase my question.

Bolzano's Thm states:
Let f(x) be continuous on [a,b].
f(x) changes sign on [a,b] => f(x) has a zero on [a,b]

I have a problem where f(x) changes sign but on the interval of all real numbers, from negative infinity to infinity. I would like to use Bolzano's Theorem, but my interval is open, not closed as worded in the theorem. Any suggestions?

**Drexel28** · May 18th, 2010, 06:35 PM

Originally Posted by mitwhiz

Let me rephrase my question.

Bolzano's Thm states:
Let f(x) be continuous on [a,b].
f(x) changes sign on [a,b] => f(x) has a zero on [a,b]

I have a problem where f(x) changes sign but on the interval of all real numbers, from negative infinity to infinity. I would like to use Bolzano's Theorem, but my interval is open, not closed as worded in the theorem. Any suggestions?

If it's continuous on $[a,b]$ and $f$ changes signs at $x_0,x_1\in (a,b)$ then what is true of $f$ on $\left[x_0-\tfrac{x_0-a}{2},x_1+\tfrac{x_1+a}{2}\right]$ ?

**mitwhiz** · May 18th, 2010, 08:01 PM

I'm not sure. Basically, I want Bolzano's Theorem to be true for (a,b), in particular (-inf, inf). Could you argue that an open interval is a subset of a closed one? Therefore it applies for an open interval too?

**Drexel28** · May 18th, 2010, 08:13 PM

Originally Posted by mitwhiz

I'm not sure. Basically, I want Bolzano's Theorem to be true for (a,b), in particular (-inf, inf). Could you argue that an open interval is a subset of a closed one? Therefore it applies for an open interval too?

My point is that if your function changes signs in the open interval you can form a closed subinterval of your open interval which contains those two points. Apply the IVT there.

**HallsofIvy** · May 19th, 2010, 04:44 AM

Originally Posted by mitwhiz

I want to use Bolzanos Theorem to prove that a function changes sign on an interval from -inf to inf. However, this theorem is defined for closed intervals, and the infinity interval is an open one.

Yes, but that does not mean it is not closed! In any topology the entire set is both open and closed.

Is there an obvious and easy way to prove this for an open interval, or in the special case of (-inf, inf)?

Yes, use the fact that (-inf, inf) is closed!

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