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Qurstion:

Show that if $\mathbf{B}$ is a basis for a topology on X then the collection $\mathbf{B}_{Y}=\left \{ B\cap Y:B \in \mathbf{B} \right \}$ is a basis for the subspace topology on Y.

$\mathbf{B}$ is a basis for a topology on X means that X is non-empty an satisfies the definition for what it means to be a basis for a topology.

Any hints is appreciated.

Thanks in advance.

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  • $\begingroup$ Why can't you use the definition to show that $B \cap Y$ is a basis for Y? Where are you getting stuck? $\endgroup$ – астон вілла олоф мэллбэрг Sep 11 '16 at 4:26
  • $\begingroup$ I am unable to show that for every element y in Y, there is a basis element in $B_{Y}$ such that y is in that basis element. $\endgroup$ – Mathematicing Sep 11 '16 at 4:39
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So we know that $B$ is a basis for the topology on $X$. To show that $B \cap Y$ is a basis for $Y$, we need to show that every open set $U \subset Y$ can be written as a union of sets of the form $B \cap Y$.

Let $U$ be an open set contained in $Y$. Then , because $Y$ is under the subspace topology of $X$ (I'm assuming this), $U$ is also an open set in $X$. Because $B$ is a basis for $X$, $U$ can be written as a union of sets from $B$, i.e.$$U = \bigcup_{\alpha \in A} B_\alpha$$ where $A$ is some indexing set and $B_\alpha \in B$ for every $\alpha$.

Now, just take the intersection with $Y$: $$ U \cap Y = U = \bigg(\bigcup_{\alpha \in A} B_\alpha\bigg) \cap Y = \bigg(\bigcup_{\alpha \in A} (B_\alpha \cap Y)\bigg) $$

That is, $U$ has been written as a union of elements of the form $B_\alpha \cap Y$, where $B_\alpha \in B$. Hence, because $U$ was an arbitrary open set, $B \cap Y$ forms a basis for the subspace topology on $Y$.

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  • $\begingroup$ Basis for Y? We are suppose to show basis for a topology on Y. $\endgroup$ – Mathematicing Sep 11 '16 at 5:46
  • $\begingroup$ That was my intention,my friend. Anyway, thank you for pointing out the mistake. $\endgroup$ – астон вілла олоф мэллбэрг Sep 11 '16 at 5:48

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