Let $\frak T$ be a totally ordered set of topologies on $\Bbb R$.
Is $|\frak T|\le |\Bbb R|$?
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Sign up to join this communityLet $\frak T$ be a totally ordered set of topologies on $\Bbb R$.
Is $|\frak T|\le |\Bbb R|$?
The answer is no. As Ashutosh notes in the comments, a modification of my original answer shows in ZFC that there is a chain longer than $|\mathbb{R}|$.
Theorem. There is a linearly order chain of topologies on $\mathbb{R}$ of size larger than $\mathbb{R}$.
Proof. Let $\kappa$ be the smallest cardinal so that $2^\kappa>|\mathbb{R}|={\frak c}=2^\omega$. It follows that the tree $T=2^{<\kappa}$ has size at most continuum, since $\kappa\leq\frak{c}$ and so $2^{<\kappa}$ is a $\kappa$-sized union of sets of size at most $\frak{c}$, and hence of size at most $\frak{c}$. Thus, we may label the nodes in $2^{<\kappa}$ with reals, and get $2^\kappa$ many paths through this tree. For each path $p$, let $X_p$ be the set of reals appearing on $p$ or to the left of $p$. So we have a linearly ordered subset of $P(\mathbb{R})$ of size $2^\kappa$, which is larger than the continuum. For each $X\subset\mathbb{R}$, let $\tau_X$ be the topology with the discrete topology on $X$ and the indiscrete topology on $\mathbb{R}-X$. Since $X\subset Y\iff \tau_X\subset \tau_Y$, the chain in $P(\mathbb{R})$ translates to a chain of topologies on $\mathbb{R}$ of size $2^\kappa$, and so we have a chain larger than $|\mathbb{R}|$, as desired. QED
So in ZFC we may find a chain of topologies longer than $|\mathbb{R}|$.
Previous answer:
The answer is no, not necessarily. In particular, if the continuum hypothesis holds (or more generally, if merely $|\mathbb{R}|<2^{\omega_1}$, which is to say that Luzin's hypothesis fails), then there is a very long chain.
Theorem. There is a linearly ordered chain of topologies on $\mathbb{R}$ of size $2^{\omega_1}$. This is strictly larger than $|\mathbb{R}|$ if the continuum hypothesis holds, or more generally, if Luzin's hypothesis fails.
Proof. Consider the tree $T=2^{<\omega_1}$ consisting of countable ordinal length binary sequences. This tree has size continuum, and so we may label each node with a real. Consider the paths through this tree, ordered from left-to-right, which is the lexical order. There are $2^{\omega_1}$ many such paths. For each path, consider the set of reals appearing on the path or to the left. This gives a linearly ordered chain in $P(\mathbb{R})$ of size $2^{\omega_1}$. For each subset $X\subset\mathbb{R}$, let $\tau_X$ be the topology consisting of the discrete topology on $X$, and the indiscrete topology on $\mathbb{R}-X$. Thus, $X\subset Y\implies \tau_X\subset \tau_Y$. So our linearly order chain in $P(\mathbb{R})$ thus transforms to a chain in the set of topologies on $\mathbb{R}$, of size $2^{\omega_1}$. QED