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I came across the following lemma:

Let $X$ be a topological space. Suppose that $\mathcal C$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x$ in $U$, there is an element $C$ of $\mathcal C$ such that $x \in C \subset U $. Then, $\mathcal C$ is a basis for the topology of $X$

From this Lemma, can we deduce that every topological space $X$ has a basis?

To answer this question, I think all I need to do is to show that the set $\mathcal C$ in the above lemma is non-empty, but it is not clear to me how to prove/disprove this.

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Every topological space $(X, \tau)$ has a basis: $\tau$. This is independent of said lemma.

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  • $\begingroup$ You can just check that $\tau$ itself verifies the definition of a basis. Let $\Omega$ be a open set. So $\{ \Omega \}\subset \tau$ and $\Omega = \bigcup \{ \Omega \}$. Done. (it is a reunion of elements in $\tau$, namely, itself) $\endgroup$ – Ivo Terek May 21 '15 at 22:25
  • $\begingroup$ I see. Now that I read your explanation, I can let $\mathcal C = \tau$ $\endgroup$ – user74261 May 21 '15 at 22:27

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