1

In Munkres, Lemma 13.1 says that topology is the collection of all unions of elements of collection of basis elements.

Now in the figure 13.3, which is an irregular curve and also an open set i.e., element of topology, how to obtain that by union of basis elements that is circles?

I just learnt about that lemma and tried to visualise it using the figure(I know why that figure is there.)

user582661
  • 77
  • 6

2 Answers2

1

In this picture, the author was trying to prove that if you choose any two elements from a Topology $\tau$, then the intersection of that two sets is also in $\tau$. He basically proved that we can define topology by the definition of basis.

Also, note that the graphs in that picture are some regions, not the curve. The lemma tells us: If we choose any open set $O \in\tau$, then $O$ is the union of some basis elements. For example, any open set (does not matter how bad you can imagine) in $R^2$ with the standard topology is the union of some open discs.

Finally, please have a look at the example $1$ and $2$ of Munkress page $78$ (2nd edition). Here he showed two different bases $\mathbb{B}$ and $\mathbb{B^1}$ of the Euclidean topology of $\mathbb{R^2}$ as the collection of circular regions(interiors of the circles) and Open rectangles.

Answer to the last question:

Yes, any element of the basis $\mathbb{B}$ is open. More importantly, the rest of the open sets can be created by the union of the basis elements of that given basis $\mathbb{B}$.

Matha Mota
  • 3,472
  • Yes, that's ok. But my question is different. Do you want to say that this type of figure can't be obtained by circles and this cannot be an open set? – user582661 Dec 26 '19 at 09:17
  • Do you mean circle or disc? Could you read the 2nd paragraph of the answer? I edited my answer. – Matha Mota Dec 26 '19 at 09:19
  • 1
    Yeah I meant disc only...it is bit clear now. – user582661 Dec 26 '19 at 09:23
  • Just one thing can that region be obtained by union of disc? – user582661 Dec 26 '19 at 09:23
  • 1
    Yes, that's correct. Again, if your basis elements are "circles" in some topology, then you can create any open "curve" just by taking the union (countable/uncountable/finite) of them. Unfortunately, I have not seen such topologies. I believe there are some topologies like that. – Matha Mota Dec 26 '19 at 09:26
  • Ok. Thank you. Can you please answer the edited question. (different from the original one) – user582661 Dec 26 '19 at 10:14
  • I can see your edited version. The last question is not clearly stated. Is the following your question? What are the open sets in $\tau$? – Matha Mota Dec 26 '19 at 10:18
  • Let us continue this discussion in chat. – Matha Mota Dec 26 '19 at 10:23
  • Yes. Basically that is my question. I have an idea about the definition given.I just want to make sure whether my implication is correct or not. – user582661 Dec 26 '19 at 10:23
  • 1
    I think we're done with the problem. Have fun with your Topology. – Matha Mota Dec 26 '19 at 10:45
  • Thank you very much. – user582661 Dec 26 '19 at 10:47
  • 1
    But in example 1,actually topology generated is same by circular regions and rectangular regions – user582661 Dec 26 '19 at 10:51
  • Sorry for the confusion. You're correct. These two are the different bases for same Euclidean Topology of the plane. Sorry I choose bad examples. I wanted to tell you: Suppose you don't have any topology rather you have a collection of sets (that will generate a topology on your space). I told you how to generate different topologies on a given set. Agian, if you know toplogy on a space there are many many bases for that topology. – Matha Mota Dec 26 '19 at 12:24
  • This thread might be helpful. https://math.stackexchange.com/questions/1852072/confusion-regarding-munkress-definition-of-basis-for-a-topology – Matha Mota Dec 26 '19 at 12:39
  • I deleted one of my comments because it was confusing and incorrect. – Matha Mota Dec 26 '19 at 12:47
1

The figure 13.3 is to illustrate the proof that $U_1,U_2 \in \mathcal{T}$, $U_1 \cap U_2 \in \mathcal{T}$. Here $U \in \mathcal{T}$ iff for each $x \in U$ we have a $B \in \mathcal{B}$ such that $x \in B \subseteq U$.

Maybe you're confused about the equivalence between "$U \in \mathcal{T}$ iff for all $x \in U$ we have a $B \in \mathcal{B}$ such that $x \in B \subseteq U$" and "$U\in \mathcal{T}$ iff $U$ is a union of elements from $\mathcal{B}$"? These are the two equivalent ways to define the topology $\mathcal{T}$ from a base $\mathcal{B}$.

See also this recent thread for some more discussion. The fact that both open squares as open discs form a base for the Euclidean plane, does indeed mean that open discs are (countable!) unions of open squares and vice versa.

Henno Brandsma
  • 242,131