# Is $[0,+\infty)$ closed and can the intermediate value theorem can be applied to functions of this domain? [closed]

The intermediate value Theorem states that if $$f:[a,b]\rightarrow\mathbb{C}$$ is a continious function then $$f$$ also must take all values in between $$f(a)$$ and $$f(b)$$.

$$\exp:\mathbb{R}\rightarrow\mathbb{C}$$ is a continious function and derived from that the functions $$\sin$$ and $$\cos$$ are also continious functions.

(Note: So far we have defined $$\cos\text{ and } \sin$$ only for $$\mathbb{R}$$)

I have a question about an application of this Theorem now.

It was used in a proof that $$\cos$$ has at least one Zero Point.

It starts with the assumption that it has no Zero Point and then uses the Argument that $$\cos$$ is continious on $$[0,+\infty]$$ + the fact that $$\cos(0)=1$$ and then makes use of the intermediate value Theorem to conclude that $$\cos$$ has no negative values. [...]

But why can we use the Theorem here, i.e why are the conditions met that we can use it?

First of all $$+\infty$$ is not even in $$\mathbb{R}$$, I must have made an mistake when I wrote down from the blackboard, we could reformulate that $$\cos$$ is continious on $$[0,+\infty)$$. But what is the Definition of this interval again?

Why is it compact? - Because otherwise we could not use the Theorem

And why does it implicate that $$\cos$$ is continious in $$\mathbb{R^+}$$?

## closed as unclear what you're asking by Lord Shark the Unknown, max_zorn, metamorphy, José Carlos Santos, CesareoJan 25 at 10:38

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• Your title poses a question that appears to be absent from your text. – Lord Shark the Unknown Jan 24 at 22:15
• It is not compact since it's not bounded – Heisenberg Jan 24 at 22:16
• $[0,\infty)$ is closed, not bounded and hence not compact in $\mathbb{R}$. – LoveTooNap29 Jan 24 at 22:18
• @LordSharktheUnknown Once a question of mine got closed because I did not put any context to it that's why I added the reason why I am asking this time. I have changed the title now. – RM777 Jan 24 at 22:19
• @Heisenberg But why can I use the intermediate value Theorem? – RM777 Jan 24 at 22:21

The cosine function is continuous on $$[0,+\infty)$$. This is not a compact interval, but it doesn't really matter. For every $$b>0$$, you can apply the intermediate value theorem on the interval $$[0,b]$$.

Suppose there exists $$b$$ such that $$\cos b<0$$, then the IVT applied to the interval $$[0,b]$$ tells us that $$\cos c=0$$, for some $$c\in(0,b)$$. Thus, if $$\cos x\ne0$$ for every $$x\in[0,+\infty)$$ we can conclude that $$\cos x>0$$, for every $$x\in[0,+\infty)$$.

The rest of the proof you are studying consists in deriving a contradiction from $$\cos x>0$$ for every $$x\ge0$$.