Bolzanos Theorem

May 2010
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0
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

MHF Hall of Honor
Nov 2009
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Berkeley, California
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 \(\displaystyle \lim_{x\to-\infty}f(x)\leqslant y\leqslant \lim_{x\to\infty}f(x)\).
 
May 2010
3
0
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

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
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 \(\displaystyle [a,b]\) and \(\displaystyle f\) changes signs at \(\displaystyle x_0,x_1\in (a,b)\) then what is true of \(\displaystyle f\) on \(\displaystyle \left[x_0-\tfrac{x_0-a}{2},x_1+\tfrac{x_1+a}{2}\right]\)?
 
May 2010
3
0
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

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
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

MHF Helper
Apr 2005
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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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