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A. A non empty countable set cannot be open.
B. For a non empty set the uncountable is necessary condition for that set to be open.
C. Uncountability is a sufficient condition for a set to be open.
D. A closed set can be uncountable or countable.

My effort: If i suppose an open set (0,1) with elements belonging to Set of rational numbers, then the set (0,1) will be countable.So there do exist an open interval which is countable.Right?

for option D, The closed set [0,1] is countable in Q but not in R.hence, i say that a closed set can be both countable and uncountable.

This is a multiple select question.

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    $\begingroup$ Presumably by "set" you mean "subset of $\mathbb{R}$", and by "open" you mean "open in the Euclidean topology on $\mathbb{R}$"? $\endgroup$ – Patrick Stevens Jan 28 '17 at 13:08
  • $\begingroup$ Rereading your question, I think you need to provide us with more context. In what topologies are you working? $\endgroup$ – Patrick Stevens Jan 28 '17 at 13:09
  • $\begingroup$ I am working with line of Real numbers. Please try to answer my question on the basis of Real analysis and not using Metric spaces. $\endgroup$ – Parul Jan 28 '17 at 13:18
  • $\begingroup$ You expect me to answer a question about real analysis without using any properties of metric spaces? $\endgroup$ – Patrick Stevens Jan 28 '17 at 13:19
  • $\begingroup$ no no, u pretty much explained in easy terms. thanks $\endgroup$ – Parul Jan 28 '17 at 13:37
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The following collection of facts should answer your question:

In $\mathbb{R}$, the set $\{1\}$ is closed and finite; the set $\mathbb{N}$ is closed and countably infinite; the set $\mathbb{R}$ is closed and uncountably infinite.

The set $[0,1)$ is not open in $\mathbb{R}$, but is uncountable; the set $\mathbb{N}$ is not open in $\mathbb{R}$, but is countably infinite; the set $\{1\}$ is not open in $\mathbb{R}$, but is finite.

In $\mathbb{R}$, $\emptyset$ is open but finite.

A nonempty open set in $\mathbb{R}$ must be uncountable, since a nonempty open set containing the point $x$ must contain some open interval $B_{\epsilon}(x)$, and open intervals are uncountable. (Proof: scale and translate to $(0,1)$ wlog, then use decimal expansions to diagonalise.)

A nonempty open set in $\mathbb{R}$ with the discrete topology need not be uncountable, or even infinite: $\{1\}$ is open in this topology.

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