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Q 1. In $\mathbb R$ with the finite complement topology if $U$ is open then $U$ is uncountable. Is it true or false? Give example to support your answer.

Q 2. Let $X$ be a set and let $p$ belongs to $X$. Define a collection of sets $T = \{ U \subset X : p \in U \mbox{ or }U =\emptyset \}$. Then $T$ is a topology on $X$. Is it true or false? Give example to support your answer.

Q 3. Let $( X, T)$ be a topological space and let $S$ be a subspace of $X$. If $D$ is closed in $X$, then $S\cap D$ is closed in $S$. Is it true or false?

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You did not finish your – Jonas Meyer Nov 30 '11 at 19:53
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It seems to be homework. What did you try? – Davide Giraudo Nov 30 '11 at 19:55
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Unless they are very related, do not ask three questions in a single post. Ask them in different posts instead. – Srivatsan Nov 30 '11 at 20:06
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Welcome to MathSE. I see that this is your first question. So I wanted to let you know a few things about MathSE. We like to know the sources of questions - if it's homework, please add the [homework] tag. People will still help, so don't worry. We also like to know what you've tried on a problem or what your thoughts are, so that the answer does not re-invent the wheel. Also, many users find questions posted in the imperative ("Show that", "Prove", "Do") unpleasant and somewhat rude. These sort of pleasantries usually result in more and better answers. Thank you! – Arturo Magidin Nov 30 '11 at 20:26

closed as too localized by Asaf Karagila, t.b., Henning Makholm, Zev Chonoles Dec 20 '11 at 16:40

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1 Answer

Q1. True for $U \neq \varnothing$. Because if $U$ is open and non empty then the complement of $U$ is finite hence the cardinality of $U$ must be continuum, i.e. uncountable. Example $U=\mathbb{R} \setminus \lbrace 1,2 \rbrace$ is uncountable.

Q2. True. Because given two open sets $U_1$ and $U_2$, then if one of them is empty, their intersection is empty. Otherwise, both must contain $p$ hence their intersection contains $p$, so their intersection is open. Similarly given any family of open sets its union is open. Finally $p \in X$ and the empty set are both open so this defines a topology.

Q3. True.

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For Q1, what about $\varnothing$? – Dylan Moreland Nov 30 '11 at 20:24
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"Q3. True by the definition of the subspace topology." -- I presume that the definition of a subspace topology specifies the open sets in the subspace topology, not the closed sets. In that case, obtaining Q3 from the definition takes a bit of a verification. – Srivatsan Nov 30 '11 at 20:27

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