<s>[INST] Prove that $\left\{(x,y)\in \mathbb R^2: \ x^2-xy+y^2-1=0\right\}$ is a closed and bounded set. 
Prove that the following set $$V=\left\{(x,y)\in \mathbb R^2: \ x^2-xy+y^2-1=0\right\}$$ is closed and bounded.

The set $V$ is closed. In fact, for any continuous function $f:\mathbb R^n\rightarrow \mathbb R$, the set $\{\textbf{x}\in\mathbb R^n: f(\textbf{x})=0\}$ is closed. I would need to have a hint for the boundedness of $V$. Thanks!
 [/INST] The equation $x^2 - x  y + y^2 = 1$ can be rewritten as follows
$$\begin{bmatrix} x\\ y\end{bmatrix}^{\top} \begin{bmatrix} 1 & -\frac 12\\ -\frac 12 & 1\end{bmatrix} \begin{bmatrix} x\\ y\end{bmatrix} = 1$$
Computing the spectral decomposition,
$$\begin{bmatrix} 1 & -\frac 12\\ -\frac 12 & 1\end{bmatrix} = \frac 12 \begin{bmatrix} 1 & -1\\ 1 & 1\end{bmatrix} \begin{bmatrix} 0.5 & 0\\ 0 & 1.5\end{bmatrix} \begin{bmatrix} 1 & 1\\ -1 & 1\end{bmatrix}$$
we then obtain the following
$$\frac{1}{2} \left( \frac{x + y}{\sqrt 2} \right)^2 + \frac{3}{2} \left( \frac{x - y}{\sqrt 2} \right)^2 = 1$$
Thus, the set is an ellipse, which is closed and bounded. The ellipse's semimajor and semiminor axes are $\sqrt 2 \approx 1.414$ and $\sqrt{\frac 23} \approx 0.816$, respectively. Plotting the ellipse,

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