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Let $f,g : (\mathbb{R};J_s)\to (\mathbb{R};J_s)$, (where $J_s$ is the usual (standard) topology on $\mathbb{R}$) be continuous.

Prove or disprove:

(a) the set $\{x\in \mathbb{R} : f(x)\le g(x)\}$ is closed,

(b) the function $h :\mathbb{R} \to \mathbb{R}$ defined as $h(x) := \min \{f(x); g(x)\}$ for $x\in\mathbb{R}$ is continuous,

(c) the function $h : \mathbb{R} \to \mathbb{R}$ defined as $h(x) := \max \{f(x); g(x)\}$ for $x\in\mathbb{R}$ is continuous.

I went through the answer here but I am looking for an answer in terms of open sets.

If $f,g$ are continuous at $a$, show that $h(x)=\max\{f(x),g(x)\}$ and $k(x)=\min\{f(x),g(x)\}$ are also continuous at $a$

I tried to attempt this question but I am stuck at the definition of continuity i.e for every open set $V$ in $Y$, $f^{-1}(V)$ is open in $X$. I don't know how to use it. Another option is to use the pointwise definition of continuity but again I am unable to go any further than that.

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RagingBull
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1 Answers1

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The definition of continuity can also be given with closed sets: for every closed set $C$ of $Y$, $f^{-1}(C)$ is closed in $X$, since a set is closed if and only if its complement is open and $f^{-1}(Y\setminus V)=X\setminus f^{-1}(V)$. Here I think to $f$ as a mapping from $X$ to $Y$.

In your case, $$ \{x\in \mathbb{R} : f(x)\le g(x)\}= \{x\in \mathbb{R} : f(x)-g(x)\le0\}= h^{-1}(-\infty,0]) $$ where $h(x)=f(x)-g(x)$. Since $h$ is continuous (why?), you're done.

For (b) and (c) you should be aware of the fact that $$ \min\{a,b\}=\frac{1}{2}(a+b-|a-b|),\qquad \max\{a,b\}=\frac{1}{2}(a+b+|a-b|) $$ and that sums and differences of continuous functions are continuous; moreover the absolute value function is continuous and the composition of continuous functions is continuous.

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