# Topology Munkres ($2^\text{ed}$) $\S 16$ Exercise $2$: Subspaces of Finer Topologies

The following theorems are well known to me:
(i) Suppose $$\tau$$ and $$\tau '$$ are two topologies on a given set $$X$$. Then, $$\tau '$$ is said to be strictly finer than $$\tau$$ if $$\tau \subsetneq \tau '$$.
(ii) Let $$\mathscr{B}$$ and $$\mathscr{B'}$$ be the bases for topologies $$\tau$$ and $$\tau '$$ respectively on the set $$X$$. Then, TFAE:
$$\hspace{25pt}$$(a)$$\tau '$$ finer than $$\tau$$
$$\hspace{25pt}$$(b) for each $$x\in X$$ and each basis element $$B\in \mathscr{B}$$ containing $$x$$, there is a basis element $$B'\in \mathscr{B'}$$ such that $$x\in B'\subset B$$.
(iii) If $$\mathscr{B}$$ is a basis for the topology on $$X$$, the the collection $$\mathscr{B}_{Y}=$${$$B\cap Y| B\in \mathscr{B}$$} is a basis for the subspace topology on $$Y$$.

Question:
Suppose $$\tau$$ and $$\tau '$$ are two topologies on $$X$$, and $$\tau '$$ is strictly finer than $$\tau$$. What can you say about the corresponding subspace topologies on a subset $$Y$$ of $$X ?$$

My Attempt:
For intuition, I considered $$\mathbb{R}$$ with the standard $$\mathbb{R}_{st}$$ and lower limit $$\mathbb{R}_{l}$$ topologies whose basic open sets are given by $$(a,b)$$ and $$[a,b)$$ respectively. Now, consider $$Y=[0,1]$$ and let $$\sigma$$ and $$\sigma '$$ be the subspace topologies on $$Y$$ corresponding to $$\mathbb{R}_{st}$$ and $$\mathbb{R}_{l}$$ respectively. Let $$\mathscr{B}_{1}$$ and $$\mathscr{B}_{2}$$ be the basis corresponding to $$\sigma$$ and $$\sigma '$$ respectively.

$$\mathscr{B}_{1}=$${$$Y, \phi, [0,b),(b',1],(a'',b'')|0 ;$$0 \leq b'<1$$ and $$0\leq a''}
$$\mathscr{B}_{2}=$${$$Y, \phi, [a,b),[b',1]|0\leq a and $$0}
Therefore, every element of $$\mathscr{B}_{1}$$ is contained in $$\mathscr{B}_{2}$$. Thus, using (ii), I conclude that $$\sigma ' \subset \sigma$$.

Can you please give me hints about my mistake and also give me a way to fix it?

• The open sets of $Y$ are those of the form $Y\cap U$ where $U$ is an open set of $X$. Hence if you increase the number of $U$s, you may increase the number of open sets of $Y$. But not necessarily. – Chrystomath May 10 at 9:46
• @Chrystomath I didn't get anything from your comment. :( – Kumar May 10 at 10:00
• @Kumar By the way, I think you have a typo. You wrote $\sigma' \subset \sigma$. Did you mean $\sigma \subset \sigma'$? – ZeroXLR May 10 at 10:19
• @ZeroXLR Ya I know my conclusion is wrong and that's why I want to fix the mistake. But No Typo. :) – Kumar May 10 at 10:30
• @Kumar See my edit explaining where you made a mistake. – ZeroXLR May 10 at 10:48

If $$\tau'$$ is strictly finer than $$\tau$$, then the subspace topology $$\sigma'$$ induced on $$Y \subseteq X$$ by $$\tau'$$ will be finer than the one, $$\sigma$$, induced by $$\tau$$. Why? Well, any open set $$V$$ in $$\sigma$$ is of the form $$U \cap Y$$ where $$U$$ is in $$\tau$$. But, $$\tau'$$ is strictly finer than $$\tau$$ i.e. $$\tau \subset \tau'$$. Hence $$U$$ is in $$\tau'$$ also; so we managed to write $$V$$ as the intersection of an open set in $$\tau'$$ with $$Y$$ i.e. $$U$$ belongs to $$\sigma'$$. Thus, $$\sigma \subseteq \sigma'$$ and not $$\sigma \subset \sigma'$$. So something definitely went wrong in your example above using $$\mathbb{R}_{st}$$ and $$\mathbb{R}_\ell$$ and $$\mathscr{B}_1$$ and $$\mathscr{B}_2$$. Let us discuss that:

Therefore, every element of $$\mathscr{B}_{1}$$ is contained in $$\mathscr{B}_{2}$$.

Strictly speaking, that is not true. But you don't need that. What is true is if $$B \in \mathscr{B}_1$$ contains $$x$$, then there is an element $$B' \in \mathscr{B}_2$$ such that $$x \in B' \subseteq B$$. This is exactly what condition (ii)(b) requires. For example, if $$B = (b', 1]$$, and $$x \in (b', 1]$$, then you can just take $$B' = [x, 1]$$ to get $$x \in [x, 1] \subseteq (b', 1]$$. Next, if $$B = (a'', b'') \ni x$$, then you can take $$B' = [x, b'')$$ to get $$x \in [x, b'') \subseteq (a'', b'')$$ again. So on and so forth. And this leads to:

Thus, using (ii), I conclude that $$\sigma ' \subset \sigma$$.

You are using (ii) in exactly the opposite manner. Given what I explained above about $$\mathscr{B}_1$$ and $$\mathscr{B}_2$$, you actually get according to (ii)(a) that $$\sigma'$$ is finer than $$\sigma$$, which is just another way of saying $$\sigma \subseteq \sigma'$$ as one should expect.

Extra Fact:

In general $$\sigma \subseteq \sigma'$$ will not be strict. You can indeed have $$\sigma = \sigma'$$ even though $$\tau \subset \tau'$$ strictly. The topology you considered, $$\mathbb{R}_\ell$$, is not necessarily the best to demonstrate this. Instead, consider the $$K$$-topology $$\mathbb{R}_K$$ on $$\mathbb{R}$$ which is also mentioned in Munkres. It has as basis all open intervals $$(a, b)$$ just like $$\mathbb{R}_{st}$$. But its basis also contains sets of the form $$(a, b) - K$$ where $$K = \{\frac{1}{n} \in \mathbb{R} \ : n \in \mathbb{Z}_+\}$$.

Note what is going on. The set $$K$$ only affects open intervals that overlap with $$[0, 1]$$. So informally the topology of $$\mathbb{R}_K$$ away from that region is exactly the same as $$\mathbb{R}_{st}$$. So if for example $$Y = [6, 10]$$ or any other subset that does not overlap with $$[0, 1]$$, its subspace topology under $$\mathbb{R}_K$$ will be the same as the subspace topology induced by the standard topology. For if you have a basis element of $$\sigma'$$ of the form $$((a, b) - K) \cap Y$$, then because $$Y$$ has no elements in common with $$K$$, we have $$((a, b) - K) \cap Y = (a, b) \cap Y$$. So basis elements of $$\sigma'$$ coincide with those of $$\sigma$$.