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I'm using the following definition of closed and open sets:

  • Set $C \subset \mathbb{R}^n$ is open if $\forall x \in C\ \exists \varepsilon > 0: U_{\varepsilon}(x) \subset C$.
  • Set $C \subset \mathbb{R}^n$ is closed if it contains all its limit points.

I don't have much idea of how to tackle this exercise. This is what I've done so far (almost nothing):

Let $x_4$ be fixed, such that $x_4 \ll 1$. Then we have that

$$0 < x_1^2 + x_2^2 + x_3^2 < \ln(1+x_4^2) \approx x_4^2$$

it follows that $\forall \varepsilon > 0: U_{\varepsilon}(x_4) \not\subset C$. So $C$ is not open.

I have the feeling that $C$ is not closed either, but I don't know how to prove that or show the opposite.

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    There's a typo at the 2nd bullet point. Set $C$ is closed. – 2017-02-24
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    @GNUSupporter Ups! I fixed it (: – 2017-02-24
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    I disagree with this statement "it follows that $∀ε>0:U_ε(x_4)⊄C∀ε>0:U_ε(x_4)⊄C.$" Regarding is C closed.... I say (0,0,0,0) is a limit point of C and it is not in C. I will leave it to you to prove that statement. – 2017-02-24
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    Your inference that $C$ is not open from your statement$0 < x_1^2 + x_2^2 + x_3^2 < \ln(1+x_4^2) \approx x_4^2$ is wrong. $C$ must be open as it is the inverse of an open set under a continuous mapping. – 2017-02-25

2 Answers 2

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Hint: consider the function $f : \mathbb{R}^4 \to \mathbb{R}$ defined by: $$ f(x_1, x_2, x_3, x_4) = \ln(1+x_4^2) - x_1^2 - x_2^2 - x_3^2. $$

Show that $f$ is continuous and see what you can conclude about $C = f^{-1}[(0, \infty)]$.

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    @Surb I admit that this way is more elegant and time-saving than mine, so I upvoted. Since this solution pops up while I was struggling with the set equalities, I kept on typing until the end. IMHO, answers ([tag:general-topology]) problems can assume some basic well-known facts like the one you mentioned. – 2017-02-25
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$C$ is open

First, make $x_4$ the subject of the inequality in the definition of $C$ (so as to make use of preimage).

\begin{array}{rrl} & x_1^2 + x_2^2 + x_3^2 <& \ln(1+x_4^2) \\ \iff& \exp(x_1^2 + x_2^2 + x_3^2) <& 1+x_4^2 \\ \iff& \exp(x_1^2 + x_2^2 + x_3^2) - 1 <& x_4^2 \\ \iff& x_4 >& \sqrt{\exp(x_1^2 + x_2^2 + x_3^2) - 1} \text{ or } \\ &x_4 <& -\sqrt{\exp(x_1^2 + x_2^2 + x_3^2) - 1} \tag1\label1 \end{array}

Let $f_1:\Bbb R^3 \to \Bbb R$ be defined as $(x_1,x_2,x_3)\mapsto-\sqrt{\exp(x_1^2 + x_2^2 + x_3^2) - 1}$ and
$f_2:\Bbb R^3 \to \Bbb R$ be defined as $(x_1,x_2,x_3)\mapsto\sqrt{\exp(x_1^2 + x_2^2 + x_3^2) - 1}$.
They are well-defined since $x_1^2 + x_2^2 + x_3^2 \ge 0$, so $\exp(x_1^2 + x_2^2 + x_3^2) \ge 1$. This allows us to take square root.
Next, observe that $f_1$ and $f_2$ are continuous on $\Bbb R^3$ since they're composition of elementary functions.
Rewrite the definition of $C$ using these two continuous functions to see the openness.

\begin{align} C =& \{( x_1, x_2, x_3, x_4) \in \mathbb{R}^4 \mid x_1^2 + x_2^2 + x_3^2 < \ln(1+x_4^2)\} \\ \stackrel{\eqref{1}}=& \{( x_1, x_2, x_3, x_4) \in \mathbb{R}^4 \mid x_4 > \sqrt{\exp(x_1^2 + x_2^2 + x_3^2) - 1} \text{ or } \\ & x_4 < -\sqrt{\exp(x_1^2 + x_2^2 + x_3^2) - 1}\} \\ =& \{( x_1, x_2, x_3, x_4) \in \mathbb{R}^4 \mid x_4 > f_2(x_1, x_2, x_3) \text{ or } x_4 < f_1(x_1, x_2, x_3)\} \\ =& \{( x_1, x_2, x_3, x_4) \in \mathbb{R}^4 \mid x_4 < f_1(x_1, x_2, x_3) \} \cup \{( x_1, x_2, x_3, x_4) \in \mathbb{R}^4 \mid x_4 > f_2(x_1, x_2, x_3) \} \\ =& \bigcup_{x_4 \in \Bbb R} \{( x_1, x_2, x_3) \in \mathbb{R}^3 \mid x_4 < f_1(x_1, x_2, x_3) \} \cup \{( x_1, x_2, x_3) \in \mathbb{R}^3 \mid x_4 > f_2(x_1, x_2, x_3) \} \\ =& \bigcup_{x_4 \in \Bbb R} f_1^{-1}((-\infty,x_4)) \cup f_2^{-1}((x_4,+\infty)) \end{align}

Since $f_1$ and $f_2$ are continuous, for each $x_4 \in \Bbb R$, the first and the second sets in the last equality are open. Hence $C$ is open.

$C$ is not closed

Expand @DougM's idea. Consider $c_n = (0,0,0,\frac1n) \in \Bbb R^4, n \in \Bbb N$. The points $c_n$ are in $C$ since $0^2+0^2+0^2 = 0 < \ln\left(1+\frac{1}{n^2}\right)$. $(c_n)$ converges to $(0,0,0,0)$ so $(0,0,0,0) \in \overline{C}$, but $(0,0,0,0) \notin C$ since $\ln(1+0^2) = 0$, so the strictly inequality in the definition of $C$ doesn't hold.