The sentence "a.s. convergence does not come from a metric, or even from a topology" comes from a remark in Durrett's probability theory textbook, after following two theorems.

Theorem 2.3.2. $Xn \rightarrow X$ in probability if and only if for every subsequence $X_{n(m)}$ there is a further subsequence $X_{n(m_k)}$ that converges almost surely to $X$.

Theorem 2.3.3. Let $y_n$ be a sequence of elements of a topological space. If every subsequence $y_{n(m)}$ has a further subsequence $y_{n(m_k)}$ that converges to $y$ then $y_n \rightarrow y$.

The remark says "Since there is a sequence of random variables that converges in probability but not a.s., it follows from Theorem 2.3.3(above theorem) that a.s. convergence does not come from a metric, or even from a topology.

I don't know how to understand this. Although I know metric is about distance and topology is about open sets, I didn't see the connection between these. And what are some implications/applications of this property?


Let {X_n} be a sequence of random variables which converges in probability but not a.s. Take any subsequence of this sequence. The subsequence also converges in probability. This implies that there is a further subsequence which converges a.s. According to Theorem 2.3.3, if there is a topology on the space of all random variables on a given probability space in which convergence of a sequence is equivalent to a.s. convergence, then the original sequence {X_n} must itself converge a.s., which is not true.


I shall first define some notions of topology and their link with metric spaces and then answer your particular question about a.s. convergence. This is long but your initial question seems to ask for the link between topology and distance.

Almost sure convergence does not come from a topology

Take any sequence $(X_{n})_{n}$ of random variables converging in probability to some random variable $X$. Then, you know that any subsequence $(X_{n_{k}})_{k}$ converges in probability to $X$. But you know that from a sequence converging in probability to $X$, you can take a subsequence that converges almost surely to $X$. In that case, you can make every subsequence of $X_{n}$ converge almost surely to $X$. But then, if almost convergence comes from a topology, theorem 2.3.3 implies that the all sequence $(X_{n})_{n}$ converges almost surely to $X$. This is a contradiction since there exists sequences of random variables converging in probability but not almost-surely.

General definitions of metric spaces and topological spaces

A metric space is a set $X$ endowed with a distance (or metric) $d$, which is a function $d:X\times X\to\Bbb R_{+}:(x,y)\mapsto d(x,y)$ respecting the following properties:

  • For all $x,y\in X$, $d(x,y)=d(y,x)$ (symmetry);
  • For all $x,y\in X$, $d(x,y)\geq 0$ and $d(x,y)=0\iff x=y$ (positivity) and
  • For all $x,y,z\in X$, $d(x,y)\le d(x,z)+d(z,y)$ (triangular inequality).

As an exercise, you can verify that $d:\Bbb R^{2}\times \Bbb R^{2}\to\Bbb R_{+}:((x_{1},y_{1}),(x_{2},y_{2})\mapsto \max\{\vert x_{1}-x_{2}\vert,\vert y_{1}-y_{2}\vert\}$ is a distance on $\Bbb R^{2}$.

A topological space is a set $X$ endowed with a collection $\mathcal{T}$ of subsets of $X$, which are called open sets, and such that:

  • $X\in\mathcal{T}$ and $\emptyset\in\mathcal{T}$;
  • For all $U,V\in\mathcal{T}$, we have $U\cap V\in\mathcal{T}$ (stability by finite intersections) and
  • For any family $\{U_{\alpha}\}_{\alpha\in A}$ of open sets (i.e. of elements of $\mathcal{T}$), their union $\cup_{\alpha\in A}U_{\alpha}\in\mathcal{T}$ where $A$ is any set of indices ($A$ may be finite, countably infinite or uncountably infinite) (stability by unions).

A collection $\mathcal{T}$ of subsets of $X$ for which the three previous properties hold is called a "topology" on $X$. The complement in $X$ of an open set $U\in\mathcal{T}$ is called a closed set. Beware that this terminology can be dangerous: a set may be open and closed or may be neither open or closed (for a given topology)!

Link between metric space and topology

When you have a metric space $X,d$, you can define the so-called "open ball" of radius $r>0$ centered at $x_{0}\in X$:

$$B_{r}(x_{0})=\{x\in X\vert d(x,x_{0})<r\}$$

The "natural" topology on a metric space is the topology generated by the open balls. What does that mean? You define the topology to be the emptyset $\emptyset$, the finite intersections of open balls and any union of open balls. Then, obviously, this respects the property of a topology and so is a topology.

An equivalent way that is more intuitive is to say that a set $U$ is open for the topology induced by the metric if for any point $x\in U$, there exists $r>0$ such that $B_{r}(x)\subset U$, that is, there exists an open ball centered at $x$ that is still contained in $U$. This is quite intuitive, I think. One of the simplest example is $\mathbb{R}$ endowed with the "standard" distance/metric: the absolute value of the difference, that is, $d(x,y)=\vert x-y\vert$ where $x,y\in\mathbb{R}$. With the previous definition, you can determine what is an open ball and then you can check that any interval $(a,b)$ is an open set ($a\le b$), any interval $[a,b]$ ($a\le b$) is a closed set (you have to prove that $(-\infty,a)\cup(b,\infty)$ is open) and that $(a,b]$ or $[a,b)$ are neither close or open.

Notion of metrizability

Now, one can take a general topological space $X,\mathcal{T}$ and one can ask whether there exists a distance on $X$ such that the topology induced by $d$ on $X$ is the same as $\mathcal{T}$. If there exists such a distance, we say that $X,\mathcal{T}$ is metrizable. For example, if I take $\mathcal{T}$ to be all the subsets of $X$, then in particular $\mathcal{T}$ contains the singleton $\{x\}$ for every $x\in X$. And, obviously, the topology generated by the singletons is the topology that contains all the subsets of $X$ since any subset is a union of all its singletons, etc. This topology is called the discrete topology. Is there a distance which induces the same topology? Here, the answer is yes: take

$$d:X\times X\to\Bbb R_{+}:(x,y)\mapsto \begin{cases} d(x,y) = 1 &\text{if } x\neq y\\d(x,y) = 0 &\text{if } x= y\end{cases}$$

If you take $r=1/2$ for a radius, you see that open balls centered at $x$ are the singletons $\{x\}$.

(One of) The initial purpose(s) of topology

Now that the link between topology and metric spaces has been done, why would one study them? Topology is a convenient setting to talk about convergence despite having less structure than a metric space. In a metric space $X,d$, we say that a sequence $x_{n}$ converges to $x$ if for any $\epsilon>0$, there exists $N_{\epsilon}$ such that for all $n\geq N_{\epsilon}$, we have $d(x_{n},x)<\epsilon$. This is equivalent to say that for any $\epsilon>0$, there exists $N_{\epsilon}$ such that for all $n\geq N_{\epsilon}$, $x_{n}\in B_{\epsilon}(x)$, that is, from a certain index $n\geq N_{\epsilon}$, the remaining terms of the sequence are all contained in a small open ball around $x$. Naturally, in a topological space $X,\mathcal{T}$, we say that a sequence $(x_{n})_{n}$ converges to $x\in X$ (for the topology $\mathcal{T}$) if and only if, for every open set $U_{x}$ that contains $x$, there exists $N_{U_{x}}$ such that for all $n\geq N_{U_{x}}$, we have $x_{n}\in U_{x}$.

These are exactly the same definitions, but obviously the second one includes the first and is more general as there exists topological spaces that are not metrizable.

It wouldn't be fair to not talk about continuity of functions while introducing topology. Actually, I said that topology was convenient for talking about convergence, but this not exactly true. There exists topological spaces where some sequences converge to more than one point. This means that "limits" are not unique anymore (wrong choice of terminology since a "limit" is unique by definition but I hope you get the point). To ensure the unicity of "limits", one has to consider particular topological spaces called "Hausdorff" spaces or "$T_{2}$-spaces" or "separated spaces".

However, we can always talk about continuity in a well-defined manner. Let $X,d_{X}$ and $Y,d_{Y}$ be two metric spaces. A function $f:X\to Y$ is continuous at $x_{0}\in X$ if and only if for all $\epsilon>0$, there exists $\delta>0$ such that for all $x\in X$ with $d_{X}(x,x_{0})<\delta$, it holds that $d_{Y}(f(x),f(x_{0}))<\epsilon$. A function is said continuous if it is continuous at each point of $X$. We can rephrase that by saying that a function $f$ is continuous if for each $x_{0}\in X$, for each open ball $B_{\epsilon}(f(x_{0}))$ (of $Y,d_{Y}$) centered at $f(x_{0})\in Y$ and of radius $\epsilon>0$, there exists an open ball $B_{\delta}(x_{0})$ (of $X,d_{X}$) centered at $x_{0}$ such that $f(B_{\delta}(x_{0}))\subset B_{\epsilon}(f(x_{0}))$, that is, for any open ball in $Y$ containing $f(x_{0})$, there exists a small open ball in $X$ containing $x_{0}$ such that the image of this open ball is contained in the open ball containing $f(x_{0})$.

Now, you see it coming. A function $f:X\to Y$ between topological spaces $X,\mathcal{T}_{X}$ and $Y,\mathcal{T}_{Y}$ is said continuous if for any open set $U_{Y}\in\mathcal{T}_{Y}$, the preimage of $U_{Y}$ by $f$ is open in $X,\mathcal{T}_{X}$, that is, $f^{-1}(U_{Y})\in\mathcal{T}_{X}$.

Importance of metrizability

Metric spaces are particular cases of topological spaces, as suggested above. The particular structure they have because of the metric structure gives them very convenient properties that do not always hold for general topological space. Knowing that a topological space is metrizable implies that this topological space possesses all the properties of a metric space, which is very convenient. Indeed, one sometimes defines a topology to serve one's needs at the time but this topology may be useful for some applications but difficult to use for other. Knowing that it is metrizable can help broaden the range of applications of this topology.

Link between convergence and topology

One can define a topology in terms of closed sets and then define the open sets as the complements of closed sets. A closed set has the particular property that any convergence sequence in a closed set converges in the closed set, that is if $(x_{n})_{n}\subset C$ is a convergent sequence included in a closed set $C$ (closed for some topological space $X,\mathcal{T}$), then if $x_{n}\to x$ where $x\in X$, it follows that $x\in C$.

Therefore, you can define the topology induced by some convergence, for example the convergence in probability, by saying that a set is closed if it contains the limit (for that particular convergence) of every sequence contained in this set. However, one should check that this does define a topology. As you can guess from the affirmation in Durrett's book, this is not always the case.


Others have already shown how to put the two theorems together but this observation, while being true, still seems somewhat mystical to me. But I think I have an explanation for what is going on here exactly and I will post it as an answer in case anyone else feels the same way. Suppose that the statement of Theorem 2.3.2 holds. Let $\Omega_m$ denote the full measure set on which a.s. convergence occurs along a subsequence of subsequence $m$. Note that there are an uncountable number of subsequences of $1,2,\ldots$. It is not necessarily true that the intersection of an uncountable number of full measure sets has also full measure (the intersection does not even need to be measurable). So then the condition of the theorem does not imply a.s. convergence along the original sequence but it implies convergence in probability.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.