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The whole question: Let $f,g:X\rightarrow \mathbb{R}$ be continuous at $a$. Supposing that for every neighborhood $V$ of $a$ exists points $x,y$ such that $f(x)

I'm stuck. I've tried to fix some $\epsilon$ and by continuity get some neighborhood of $a$ and so get the points $x,y$ given by hyphotesis. But i couldn't work with them succesfully. I also tried the contrapositive, but i couldn't go much further.

Any hints on how to solve this problem?

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    Is $X$ an arbitrary topological space? – 2017-02-06
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    no, it's a subset of R – 2017-02-06

4 Answers 4

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Let $ h(x) = f(x) - g(x)$. Consider intervals $(a -1/n ,a + 1/n)$ around a . Thus by intermediate value theorem on the intervals we get a sequence of $ b_n$ such that $h(b_n) = 0$. Now the intersection of these intervals is just one point a and $ b_n \to a$, thus $h(a) = 0 \Rightarrow f(a) = g(a)$

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    You assume that $X$ contains an interval of the form $(a - 1/n,a+1/n)$. Also, applying the intermediate value theorem requires that $f$ be continuous over the whole interval. – 2017-02-06
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    Continuous at a point means it is continuous in a small neighbourhood and there you can have the intervals in your desired neighbourhood after some stage. – 2017-02-06
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    That is *not* what continuous at a point means. For example, the function $f:\Bbb R \to \Bbb R$ given by $$ f(x) = \begin{cases} x & x \in \Bbb Q\\ 0 & x \notin \Bbb Q \end{cases} $$ is continuous at $0$ but on no neighborhood of $0$. – 2017-02-06
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Consider the continuous function $h(x) = f(x) - g(x)$ over $X$. We note that $\lim_{x \to a}h(x) = h(a)$, call this limit $L$. Every neighborhood of $a$ contains an $x,y$ such that $h(x) < 0 < h(y)$. However, by the definition of a limit: for any $\epsilon > 0$, we can select a neighborhood $V$ such that the $x,y \in V$ must also satisfy $|h(x) - L|< \epsilon$ and $|h(y) - L| < \epsilon$.

Thus, for every $\epsilon > 0$, we have $-\epsilon < L < \epsilon$.

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    What do you mean by we can select neighborhoods such that $|h(x)-L|<\epsilon$ and $|h(y)-L|<\epsilon$? Ain't $h(x),h(y)$ just numbers for fixed points $x,y$? – 2017-02-06
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    @math.h sure. But, **every** neighborhood $V$ contains such fixed points, and as the neighborhoods get "smaller", the definition of a limit applies. – 2017-02-06
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Proposition: Let $f:X\rightarrow \mathbb{R}$ continuous at $a$. If for every neighborhood of $a$ there exists $x,y$ s.t. $f(x)>0$ and $f(y)<0$, then $f(a)=0$.

Proof: We prove the contrapositive. Supposing wlog that $f(a)<0$, since $f$ is continuous, there exists some neighborhood such that every point in it has the same signal.

Define $h:X\rightarrow \mathbb{R}$ by $h(x)=f(x)-g(x)$. Since $f,g$ are continuous, $h$ is also continuous. Let $V$ be a neighborhood of $a$ containing points $m,n$ such that $f(m)0$. Therefore, by our proposition, we say that $h(a)=0$, that is, $f(a)=g(a)$.

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Assume for a contradiction that $f(a)\neq g(a).$ WLOG, assume that $f(a)>g(a)$.

Let $d=f(a)-g(a)$.

$\exists \delta_1$ such that $|x-a|<\delta_1 \implies |f(x)-f(a)|<\frac d2$.

$\exists \delta_2$ such that $|x-a|<\delta_2 \implies |g(x)-g(a)|< \frac d2$.

Let $\delta = \min \{ \delta_1 , \delta_2 \}$.

So, $|x-a|<\delta \implies |f(x)-f(a)-g(x)+g(a)| \leq |f(x)-f(a)|+|g(x)-g(a)|<\frac d2 +\frac d2 =d$.

So, $|x-a|<\delta \implies |f(x)-g(x)-d|

So, $|x-a|<\delta \implies -d

So, $|x-a|<\delta \implies f(x)-g(x)>0$.

Thus, on a $\delta-$neighbourhood (call it $V$) of $a$, $\ f(x)>g(x)$ holds always.

Contradiction.