$X=${(x,y)$\in \mathbb R^2: 2x^2+3y^2=1$}. Endow $\mathbb R^2$ with discrete topology, and $X$ with the subspace topology. Let $X={(x,y)\in \mathbb R^2: 2x^2+3y^2=1}$. Endow $\mathbb R^2$ with discrete topology, and $X$ with the subspace topology. Then: > > A. $X$ is a compact subspace of $\mathbb R^2$ in this topology >> >> B. $X$ is a connected subspace of $\mathbb R^2$ in this topology >> >> C. $X$ is open subspace of $\mathbb R^2$ in this topology >> >> D. None of this. As singleton compact in $\mathbb R$ and f(x,y)=$2x^2+3y^2$ is continuous, so $X$ is compact. Is this right? So A is correct option?

You need to be careful with applying a polynomial saying it is continuous - the topology (and thus the notion of continuity) is very different from the usual one. It is a good exercise to try and show that a continuous map from a discrete space to $\mathbb{R}$ (with the usual topology) must be constant.

As for the exercise, the topology on $X$ is the discrete topology - Indeed, one needs to show that every singleton subset of $X$ is open in $X$, but it is already open in $\mathbb{R}^2$. (and $X$ is infinite)

Now, $X$ can't be compact - we can take all singletons as an open cover, which will have not finite subcover. Furthermore, $X$ cannot be connected - all singletons are both open and closed (and $X$ has more than one point)

We arrive at C - $X$ is open in $\mathbb{R}^2$ because every subset is - this is exactly the definition of the discrete topology : all sets are open.