$X=${(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?
 A: 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.
A: For B: If you know the definition of connectedness this should be really easy with the discrete toplogy.
For C: Even more so
A: Define a metric $d$ on $X$ which is the discrete metric.
 So,$(X,d)$ maintains its discrete topology.
But,$X$ is not totally bounded with respect to the metric $d$.
So,$X$ is not compact.
