Let f:R→Rf : \mathbb{R} \rightarrow \mathbb{R} be continuous, such that limx→∞f(x)=+∞\lim_{x \rightarrow \infty}f(x) = +\infty and limx→−∞f(x)=−∞\lim_{x \rightarrow -\infty}f(x) = -\infty. Show that f(\mathbb{R}) = \mathbb{R}f(\mathbb{R}) = \mathbb{R}

Do we begin by showing 0 is contained, and for every xx, there is -x-x and an x_0x_0 > xx?

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The intermediate value theorem should come to mind.

– Clement C.

2 days ago

2

Well, let a be in R. a is not an upper bound nor a lower bound so there exist x,yx,y so that f(x) < a < f(y)f(x) < a < f(y) so by intermediate value theorem.... Are are you expected to prove the intermediate value theorem from definitions?
– fleablood
2 days ago
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2 Answers
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Consider the function g(x)=f(x)-ag(x)=f(x)-a, \forall a \in \mathbb{R}\forall a \in \mathbb{R}.
Snce f(x)f(x) is continuous also g(x)g(x) is continuous and, from the limits for f(x)f(x), we have: \lim_{x\to -\infty}g(x)=-\infty\lim_{x\to -\infty}g(x)=-\infty and \lim_{x\to +\infty}g(x)=+\infty\lim_{x\to +\infty}g(x)=+\infty.
So there exists M>0 M>0 such that g(M)>0g(M)>0 and g(-M)<0g(-M)<0. Using the intermedite value theorem in the interval [-M,M][-M,M] we find that there is a point x_0x_0 such that
g(x_0)=f(x_0)-a=0
g(x_0)=f(x_0)-a=0
so there is a value x_0x_0 such that f(x_0)=a \quad \forall a \in \mathbb{R}f(x_0)=a \quad \forall a \in \mathbb{R}
There will always be, for arbitrary N\in\Bbb NN\in\Bbb N, points a<0

Thus [-N,N]\subset [f(a),f(b)]\subset f([a,b])\subset f(\Bbb R)[-N,N]\subset [f(a),f(b)]\subset f([a,b])\subset f(\Bbb R) by the intermediate value theorem.