# Munkres Topology Theorem 23.3 Proof Clarification & Clarifications Regarding Subspaces

Munkres provides us with this lemma before performing the proof.

This proof is very clear; then, we are presented with the following theorem, with which I am having some difficulties.

My first question is, what topology do the sets $$C$$ and $$D$$ belong to? We have not assumed that $$Y$$ is a subspace, only that the sets in the union are connected subspaces; is it just assumed that we are taking $$Y$$ as a subspace of $$X$$?

Similarly, my second question is with respect to the highlighted area. I don't see how this follows from the lemma; $$A_\alpha$$ has not been specified as being taken in the subspace topology of $$Y$$, yet the use of the lemma would require $$A_\alpha$$ be a subspace of $$Y$$.

I think, perhaps, I have a fundamental misunderstanding of when we are permitted to take subsets as being in the subspace topology. Can we just do it arbitrarily, and have it be implied we are considering certain subsets as subspaces?

• Any separation of Y will lead to a separation of $A_\alpha$ and any subset of a topological can be given subspace by intersection of open sets of parent space with the subset of concern. – Tutankhamun Oct 20 '18 at 1:35
• This I understand. In the proof of the theorem, are we implicitly taking $C,D$ as open sets in $X$, meaning $Y$ is considered as subspace of $X$? Then, are we taking the $A_\alpha$ as subspaces of $Y$? – Saru Oct 20 '18 at 1:45
• C and D are separation of Y they must be open subsets of Y. Yes, we are taking them as subspaces – Tutankhamun Oct 20 '18 at 1:48
• Ah yes! This makes much more sense to me now. Thanks! – Saru Oct 20 '18 at 1:53
• Every subset of $X$ is given the subspace topology as a subspace of $X$ in this theorem. – DanielWainfleet Oct 20 '18 at 7:45

Is it just assumed we are taking $$Y$$ as a subspace of $$X$$?

The context says yes, especially since $$Y$$ was just introduced as such in the Lemma you cited.

So, in this Theorem, $$C$$ and $$D$$ are open in $$Y$$ in the subspace topology.

I don’t see how this follows from the Lemma.

Let $$Y$$ be your “$$X$$” and $$A_{\alpha}$$ be your “$$Y$$”, then apply the Lemma. That’s it.

...require $$A_{\alpha}$$ to be a subspace of Y

Yup!

Can we just do it arbitrarily?

Pretty much. Recall that the subspace topology on any subset $$Y$$ of $$X$$ has basis of the form $$Y \cap U$$, where $$U$$ is open in $$X$$.

The open $$U$$’s are what give us the topological structure; the $$Y$$ they are attached to is irrelevant in that respect. The $$Y$$ can be anything—other than the empty set, of course.

• This was really helpful. So, essentially, $A_\alpha$ are compact subspaces of $X$, and we're considering $Y=\cup A_\alpha$ in the subspace topology. Then, we suppose $Y$ is separable. Then, since $A_\alpha\subset Y$, we consider $A_\alpha$ in the subspace topology of $Y$. Since $Y$ is separable and $A_\alpha$ is connected subspace of $X$, we have $A_\alpha$ is connected subspace of $Y$. Thus, $A_\alpha$ lies in either $C$ or $D$, but not both. – Saru Oct 20 '18 at 2:07

A space $$X$$ is connected iff we cannot write it as the union of two disjoint open non-empty subsets (such a partition is called a separation (or disconnection)). A subset $$A$$ of a space is connected iff it is connected in the subspace topology. If we talk about $$A \subseteq X$$ having some (topological) property, by default we mean that $$A$$ has that property as a subspace of $$X$$, in the subspace topology.

So $$Y$$ in your example has the subspace topology wrt $$X$$ and $$C$$ and $$D$$ are open in $$Y$$ (being a separation). The lemma applies in that space $$Y$$ as total space, with each $$A_\alpha$$ as the connected subspace. Subspace topologies are transitive, so the subspace topology of $$A_\alpha$$ wrt $$X$$ is the same as wrt $$Y$$. So the lemma applies.