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 $Y$.

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.

$\mathscr{B}_{1}=${$Y, \phi, [0,b),(b',1],(a'',b'')|0<b\leq1$ ;$0 \leq b'<1$ and $0\leq a''<b''\leq1 $}
$\mathscr{B}_{2}=${$Y, \phi, [a,b),[b',1]|0\leq a<b\leq1$ and $0<b'\leq 1$}
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?

  • $\begingroup$ 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. $\endgroup$ May 10, 2019 at 9:46
  • 1
    $\begingroup$ @Chrystomath I didn't get anything from your comment. :( $\endgroup$
    – Kumar
    May 10, 2019 at 10:00
  • $\begingroup$ @Kumar By the way, I think you have a typo. You wrote $\sigma' \subset \sigma$. Did you mean $\sigma \subset \sigma'$? $\endgroup$
    – balddraz
    May 10, 2019 at 10:19
  • $\begingroup$ @ZeroXLR Ya I know my conclusion is wrong and that's why I want to fix the mistake. But No Typo. :) $\endgroup$
    – Kumar
    May 10, 2019 at 10:30
  • 1
    $\begingroup$ @Kumar See my edit explaining where you made a mistake. $\endgroup$
    – balddraz
    May 10, 2019 at 10:48

1 Answer 1


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 $\sigma$.


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