Closedness of the set $A = \{x \in X \mid f(x) \geq g(x)\}$

Question

Let $(X, \Vert\cdot\Vert_X)$ be a normed space and the functions $f, g : X \to \mathbb R$ continuous. Show that the set $A = \{x \in X \mid f(x) \geq g(x)\}$ is closed.

A failed attempt

We should show that $f(x) \geq g(x)$, when $x$ is an arbitrary accumulation point of $A$.

The functions $f$ and $g$ are continuous, so for every $\varepsilon_1> 0$ there is a $\delta_1 > 0$, for which $$ \Vert x - x_n \Vert_X < \delta_1 \quad\Rightarrow\quad \Vert f(x) - f(x_n) \Vert_Y < \varepsilon_1, $$ where now $x$ is an accumulation point of $A$ and $x_n$ a member of a sequence $(x_n) \subset A : x_n \to x$ as $n \to \infty$. The same continuity condition holds for the function $g$, possibly with different $\varepsilon_2$ and $\delta_2$.

We also know, that for a member of the sequence $(x_n) \subset A$, $f(x_n) \geq g(x_n)$. Intuitively, as we then approach the accumulation point $x$ from inside the set and the functions are continuous, $f$ should stay above $g$ as we hit $x$.

But how do I state this mathematically? Should I observe how the difference $\Vert f(x) - g(x_n)\Vert$ is always greater than the difference $\Vert f(x) - f(x_n)\Vert_Y$ for all $x_n$ somehow? My guess is that I should show that I can choose a $\delta$ (maybe $\Vert f(x_n) - g(x_n)\Vert_Y$), but is that really the direction I should take?

I am especially interested in arguments based on the definitions of closedness in terms of accumulation points and continuity.

Answer 1

The function $h=f-g$ is continuous and$$A=h^{-1}\bigl([0,\infty)\bigr).$$Since $[0,\infty)$ is closed, $A$ is closed too.

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Here's another approach, which proves that $A^\complement$ is an open set. Take $x\in A^\complement$. Then $h(x)<0$. Therefore, since $h$ is continuous, there is a $r>0$ such that$$\lVert y-x\rVert<r\implies\bigl\lvert h(y)-h(x)\bigr\rvert<\bigl\lvert h(x)\bigr\rvert\implies h(y)<0.$$But then $B_r(x)\subset A^\complement$. Since, for each $x\in A^\complement$, there is an open ball centered at $x$ and contained in $A^\complement$, $A^\complement$ is an open set.

Answer 2

Fix $\epsilon>0$. Then for sufficiently large $n$, we have $|f(x)-f(x_n)|<\epsilon$ and $|g(x)-g(x_n)|<\epsilon$, so $$f(x)-g(x)=(f(x)-f(x_n))+(f(x_n)-g(x_n))+(g(x_n)-g(x)) \geq -2\epsilon.$$ Notice the left hand side does not depend on $\epsilon$. So we have $f(x)-g(x) \geq -2\epsilon$ for all $\epsilon>0$, which implies $f(x)-g(x) \geq 0$.

Answer 3

$h(x) =f(x)-g(x)$, $x \in X$ ;

Show that $A^c=\{h(x) <0| x \in X\}$ is open.

Let $x_0 \in A^c $, then $h(x_0) <0$.

Since $h$ continuos:

$\epsilon >0$ given, there is a $\delta$ s.t.

$||x-x_0|| <\delta$ implies $|h(x)-h(x_0)| <\epsilon$.

$-\epsilon +h(x_0) <h(x)<\epsilon +h(x_0)$.

Choose $\epsilon =-h(x_0)/2 >0$.

Then:

There is a $\delta_0$ s.t. $||x-x_0|| <\delta_0$ implies

$h(x) <\epsilon + h(x_0) = h(x_0)/2<0$, i.e.

$B_{\delta_0}(x_0) \subset A^c$.

$A^c$ is open, hence $A$ is closed.