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I saw the following claim in this thread :

For a function with the intermediate-value property the left- and right-handed limits at $x$, if they exist, equal $f(x)$.

A function $f: \mathbb{R} \to \mathbb{R}$ is said to have the intermediate-value property if for any $a$,$b$ and $\lambda \in [f(a),f(b)]$ there is a $x \in [a,b]$ such that $f(x)=\lambda.$

A failed attempt:

Let $f$ be a real function on $\mathbb{R}$ with the intermediate property. Let $p$ be a point where the function possesses left hand and right hand limits $f(p)_-, f(p)_+.$

To begin with I should be able to show that $f(p)_- = f(p)_+=l_p$(which I am not able to show)

Once that is shown, I tried this . . .

let $(x_n)$ be a sequence in $(-\infty,p)$ converging to $p$ and $(y_n)$ be a sequence in $(p,\infty)$ converging to $p.$

Then $x_n < y_n$ for each $n\in \mathbb{N}.$

For each $n$ let $\lambda_n\in \mathbb{R}$ be such that $\lambda_n \in [f(x_n),f(y_n)]. $ By the intermediate value property there exists $p_n \in (x_n,y_n)$ such that $f(p_n)=\lambda_n.$

It is evident using sandwich theorem that $p_n$ converges to $p$ and $\lambda_n=f(p_n)$ converges to $l_p$

. . . and this is leading nowhere

Can anyone show me a way to do this?

Bijesh K.S
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  • By existence of left right limits at $p$ it is ensured that values to the left of $p$ are near $f(p-) $ and values to the right of $p$ are near $f(p+) $. If $f(p-) \neq f(p+) $ then there will be some values between these two numbers which are not taken by $f$ and that violates intermediate value property. – Paramanand Singh Jan 06 '18 at 09:34

1 Answers1

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Assume $f(p)_- \neq f(p)_+$, wlog $f(p)_- < f(p)_+$. Let $c:=f(p)_+ - f(p)_-$

Then by definition of the one-sided limit there is $\varepsilon > 0$ such that

$$f(x)<f(p)_- + \frac{c}{4}\, \text{ if } x\in (p-\varepsilon, p)$$ and $$ f(x) > f(p)_+ - \frac{c}{4} \, \text{ if } x\in (p, p+\varepsilon)$$

It should now be easy to verify that this will imply a contradiction to the intermediate value property.

Thomas
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  • Okay so this means that if $\lambda \in [f(p)- + \frac{c}{4},f(p)+ - \frac{c}{4}]\subseteq [f(x),f(y)]$ then there is no $c\in [x,y]$ such that $f(c)=\lambda.$ – Bijesh K.S Jan 06 '18 at 11:53
  • $x \in (p-\epsilon,p)$ and $y\in (p,p+\epsilon)$ – Bijesh K.S Jan 06 '18 at 12:09
  • and in a similar way I should be able to prove $f(p)=f(p)- = f(p)+$ – Bijesh K.S Jan 06 '18 at 12:21
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    @BijeshK.S well, almost. There is still $p$ around, which will allow you to catch one of the values -- $f(p)$ might hit, e.g., $f(p)- + \frac{c}{2}$ -- but then you will miss $f(p)- + \frac{5c}{8}$, say. That's nitpicking, of course, but if you want to be strict you must not ignore this point. – Thomas Jan 06 '18 at 13:02