Part of the definition of a basis for a topology for a set $X$ states:

If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$, then there is a basis element $B_3$ containing $x$ such that $B_3 \subset B_1 \cap B_2$.

I would like to clarify that this definition does not rule out the possibility that $B_3=B_2$ or $B_3=B_1$. Because applying this definition to, for instance, $B_1$ and $B_3$ we should find a set $B_4$ containing $x$ such that $B_4 \subset B_1 \cap B_3$.

I have another question about basis. Is it possible for a basis on $X$ to be also a topology on $X$? For instance for set $X=\{1,2\}$, basis $\mathcal{B} = \{ \phi, \{1\}, \{1,2\}\}$ is also a topology (I hope I am applying the definitions correctly). If $\mathcal{B}$ it is indeed a basis and topology for this particular example, then I find this a bit counter-intuitive because one expects a basis to be smaller (properly contained) than a topology.

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    $\begingroup$ Basis need not be smaller. Any topology is a basis for itself. $\endgroup$ – Kavi Rama Murthy Oct 27 '20 at 6:41

As pointed out in the comment, every topology is its own basis. But every basis need not be a topology.

The proof of every topology being a basis is quite simple.

  1. For every element $x\in X$, we have an element in the topology which contains $x$. This element is $X$ itself.
  2. For two open sets $X_1,X_2,X_1\cap X_2$ is open and belongs to the topology as well.

Thus both properties of a basis are satisfied.

  • $\begingroup$ Thank you for clarification! Is my observation in the first half of the question correct? That it is possible that $B_3 = B_2$ or $B_3 = B_1$. $\endgroup$ – Rob Oct 27 '20 at 7:12
  • $\begingroup$ Yes, it is true. $\endgroup$ – Shubham Johri Oct 27 '20 at 7:14
  • $\begingroup$ @Rob If my answer was satisfactory, consider accepting it by clicking the tick mark button next to it. $\endgroup$ – Shubham Johri Oct 28 '20 at 11:28

Any topological space is open set. That make the definition of the topologial space.

If you'll see then you find that basis elements are nothing but open sets. If $B_1$ and $B_2$ are basis of the topological space $X_\tau$. Then there is a basis $B_3$ such that $B_3\subset B_1\cap B_2$. In order to visualize it draw two circles and intersect them you can find an another circle in intersection area. Whatever is your question think about in this way. Hope It'll full fill your desire.


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