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I would like someone to check my working please. Here is the question:

Is the mapping

[tex]f:\mathbb{Q}_p \rightarrow \mathbb{R}[/tex]

continuous?

My solution:

A mapping [itex]f:X\rightarrow Y[/itex] is continuous at a point [itex]x \in X[/itex] if for every [itex]\epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that

[tex]d(x,y) < \delta \Rightarrow d(f(x),f(y)) < \epsilon[/tex]

A function is continuous if it is continuous at every point in its domain.

This is the definition of continuity that I will be using. Note that I made this choice because the domain of the mapping lies in the rationals with respect to the p-adic metric. Further, [itex]\mathbb{Q}_p[/itex] and [itex]\mathbb{R}[/itex] are both metric spaces - ie there is a very good notion of distance between two points in both sets. This idea is crucial to my solution.

Loosely, the definition above says that all points [itex]f(y) \in Y[/itex] near [itex]f(x) \in Y[/itex] have their preimages [itex]y \in X[/itex] near [itex]x \in X[/itex].

In the question at hand, the function [itex]f[/itex] maps points in the metric space [itex]\mathbb{Q}_p[/itex] to points in [itex]\mathbb{R}[/itex] - where [itex]\mathbb{R}[/itex] is the metric space w.r.t to usual Euclidean metric.

For the function to be continuous the following must be satisfied: The point [itex]f(x) \in \mathbb{R}[/itex] near [itex]f(x) \in \mathbb{R}[/itex] must imply that [itex]y \in \mathbb{Q}_p[/itex] is near [itex]x \in \mathbb{Q}_p[/itex].

Technically speaking, given an epsilon > 0 there must exist a delta > 0 such that [itex]d(x,y) < \delta \Rightarrow d(f(x),f(y)) < \epsilon[/itex].

But [itex]d(x,y) := |x-y|_p[/itex]. That is, distance between two points with respect to the p-adic metric gets larger and larger as [itex]x[/itex] gets closer and closer to [itex]y[/itex]

To make this clearer, the distance between two points [itex]x, y \in \mathbb{Q}_p[/itex] actually gets larger as [itex]x[/itex] and [itex]y[/itex] become closer with respect to the Euclidean metric. So the preimages of 'close' points in [itex]\mathbb{R}[/itex] are actually very far apart in [itex]\mathbb{Q}_p[/itex]. This is the exact opposite of what it means for a function to be continuous.

How does this sound?

Is the mapping

[tex]f:\mathbb{Q}_p \rightarrow \mathbb{R}[/tex]

continuous?

My solution:

A mapping [itex]f:X\rightarrow Y[/itex] is continuous at a point [itex]x \in X[/itex] if for every [itex]\epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that

[tex]d(x,y) < \delta \Rightarrow d(f(x),f(y)) < \epsilon[/tex]

A function is continuous if it is continuous at every point in its domain.

This is the definition of continuity that I will be using. Note that I made this choice because the domain of the mapping lies in the rationals with respect to the p-adic metric. Further, [itex]\mathbb{Q}_p[/itex] and [itex]\mathbb{R}[/itex] are both metric spaces - ie there is a very good notion of distance between two points in both sets. This idea is crucial to my solution.

Loosely, the definition above says that all points [itex]f(y) \in Y[/itex] near [itex]f(x) \in Y[/itex] have their preimages [itex]y \in X[/itex] near [itex]x \in X[/itex].

In the question at hand, the function [itex]f[/itex] maps points in the metric space [itex]\mathbb{Q}_p[/itex] to points in [itex]\mathbb{R}[/itex] - where [itex]\mathbb{R}[/itex] is the metric space w.r.t to usual Euclidean metric.

For the function to be continuous the following must be satisfied: The point [itex]f(x) \in \mathbb{R}[/itex] near [itex]f(x) \in \mathbb{R}[/itex] must imply that [itex]y \in \mathbb{Q}_p[/itex] is near [itex]x \in \mathbb{Q}_p[/itex].

Technically speaking, given an epsilon > 0 there must exist a delta > 0 such that [itex]d(x,y) < \delta \Rightarrow d(f(x),f(y)) < \epsilon[/itex].

But [itex]d(x,y) := |x-y|_p[/itex]. That is, distance between two points with respect to the p-adic metric gets larger and larger as [itex]x[/itex] gets closer and closer to [itex]y[/itex]

**with respect to the Euclidean metric**. So I can not always guarantee the existance of a positive, non-zero [itex]\delta[/itex] for any given [itex]\epsilon[/itex]. Therefore the mapping [itex]f[/itex] is NOT continuous.To make this clearer, the distance between two points [itex]x, y \in \mathbb{Q}_p[/itex] actually gets larger as [itex]x[/itex] and [itex]y[/itex] become closer with respect to the Euclidean metric. So the preimages of 'close' points in [itex]\mathbb{R}[/itex] are actually very far apart in [itex]\mathbb{Q}_p[/itex]. This is the exact opposite of what it means for a function to be continuous.

How does this sound?

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