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Let $X=\left\{(x, y) \in \mathbb{R}^{2}: 2x^{2}+3y^{2}= 1\right\}$. Endow $\mathbb{R}^{2}$ with the discrete topology, and $X$ with the subspace topology. Then.

  1. $X$ is a compact subset of $\mathbb{R}^{2}$ in this topology.
  2. $X$ is a connected subset of $\mathbb{R}^{2}$ in this topology.
  3. $X$ is an open subset of $\mathbb{R}^{2}$ in this topology.
  4. None of the above.
in Linear Algebra by Boss (30.2k points) | 84 views

1 Answer

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A subspace of discrete topological space is discrete hence X is discrete top. space.

Now one can easily see the ellipse represented by X in R^2.

A. It is a uncountable set with discrete top. so can't be compact(because we can take open cover consisting singleton set).

B. X is not connected take any element of X and it's complement in X they form a separation of X.

C. In discrete top. each subset is open.

So ans is C.
by (11 points)

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