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For a function $f: \mathbb{R}^n \to \mathbb{R}$, I was wondering

  1. Is $\{x \in \mathbb{R}^n: f(x) < 0\}$ open?

    If not, what are some sufficient and/or necessary conditions for it to be open?

  2. Is $\{x \in \mathbb{R}^n: f(x) = 0\}$ closed?

    If not, what are some sufficient and/or necessary conditions for it to be closed?

  3. Is $\{x \in \mathbb{R}^n: f(x) \leq 0\}$ closed?

    If not, what are some sufficient and/or necessary conditions for it to be closed?

  4. I feel $\{(x,y) \in \mathbb{R}^2: -x+\sqrt{y} < c \}$ for some $c \in \mathbb{R}$ is open. To prove it, I want to show that for every point in it, it has an open ball centered at it and the ball is contained inside the subset. This is however not obvious for me to show.

Thanks and regards!

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Are you assuming that $f$ is continuous? Then (1)-(3) are true. You expression in (4) is not defined everywhere, so that's slightly awkward. – Dylan Moreland Feb 7 '12 at 14:36
@DylanMoreland: Thanks! On the contrary, all the questions are for proving some other function $g$ to be continuous. For every point $x$ in the domain of $g$, I started by taking a small open ball around $g(x)$ with radius $\epsilon$, and want to show there exists a open neighborhood of $x$ whose image under $g$ is contained in the open ball around $g(x)$. I want to find the open neighborhood of $x$ by solving the inequality $g(x) < \epsilon$ – Tim Feb 7 '12 at 14:38
They certainly aren't true for just any function. I could define $f$ by sending $0$ to $-1$ and everything else to $0$, and that would ruin (1). What function are you trying to prove continuous? – Dylan Moreland Feb 7 '12 at 14:41
@DylanMoreland: $g([x,y])=-x+\sqrt{y}$ when $x \leq \sqrt{y}$, and $0$ elsewhere. – Tim Feb 7 '12 at 14:51
Are you assuming that $g(x) = 0$? It seems to me that you'd want to find $y \in \mathbf R^n$ such that $-\epsilon < g(x) - g(y) < \epsilon$. – Dylan Moreland Feb 7 '12 at 14:52
up vote 6 down vote accepted

In general there is no such reasonable conditions. For cases (1)-(3) consider functions $$ f_1(x)=\begin{cases}1\qquad x\neq(0,\ldots,0)\\-1\qquad x=(0,\ldots,0)\end{cases} $$ $$ f_2(x)=\begin{cases}0\qquad x\neq(0,\ldots,0)\\1\qquad x=(0,\ldots,0)\end{cases} $$ $$ f_3(x)=\begin{cases}0\qquad x\neq(0,\ldots,0)\\1\qquad x=(0,\ldots,0)\end{cases} $$ respectively. In this cases the set $$ \{x\in\mathbb{R}^n:f_1(x)<0\}=\{(0,\ldots,0)\} $$ is not open, the set $$ \{x\in\mathbb{R}^n:f_2(x)=0\}=\{x\in\mathbb{R}^n:x\neq(0,\ldots,0)\} $$ is not closed and the set $$ \{x\in\mathbb{R}^n:f_3(x)\leq 0\}=\{x\in\mathbb{R}^n:x\neq(0,\ldots,0)\} $$ is not closed also. But if require $f$ to be continuous then all these statements will be true. In fact the following general result holds.

Theorem. Let $X$, $Y$ be metric spaces and $f:X\to Y$ map between them, then conditions

  1. $f$ is continuous

  2. for each open set $U\subset Y$ the set $f^{-1}(U)\subset X$ is open

  3. for each closed set $F\subset Y$ the set $f^{-1}(F)\subset X$ is closed

are equivalent.

Now you can apply this result to the continuous function $$ f(x,y)=-\sqrt{x}+y $$ where $X=\mathbb{R}_+\times\mathbb{R}$, $Y=\mathbb{R}$ and $U=(-\infty,c)$.

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