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Question: There is a function ##f##, it is given that for every monotonic sequence ##(x_n) \to x_0##, where ##x_n, x_0 \in dom(f)##, implies ##f(x_n) \to f(x_0)##. Prove that ##f## is continuous at ##x_0##

Proof: Assume that ##f## is discontinuous at ##x_0##. That means for any sequence ##(x_n)##, in ##dom(f)##, converging to ##x_0## we don't have ## \lim f(x_n) = f(x_0)##, symbolically

For every N there exists, at least one, ##k > N## such that

$$

|x_k - x_0| < \delta ~ \text{but} ~ |f(x_k) f(x_0)| \geq \varepsilon

$$

Doubt 1: I haven't said anything about ##\delta## and ##\varepsilon## before, and simply introduced them to mean the difference is small,is this how we do it formally?

Now, as for every ##N## we have "a" ##k## satisfying those properties, therefore, we have all those ##k## forming a subset of ##\mathcal{N}## like thus

$$

k_1 > N_1 ~~\text{satisfying those properrties}$$

$$k_2 > N_2 > k_1 ~~ \text{satisfying those properties}$$

##\vdots ##

$$k_i > N_i > k_{i-1}$$

##\vdots##

and hence, ##(x_k)## is a sub-sequence of ##(x_n)##, and

$$\lim (x_k) \to x_0$$

##\since## for every ##x_k## we have ##|f(x_k) - f(x_0)| \geq \varepsilon##, but ##(x_k)## is a monotone sequence and we must have ##|f(x_k) - f(x_0)| < \varepsilon##

This contradiction has arisen because our assumption that ##f## was discontinuous was wrong.

Doubt #2: Is what I have done correct?

Proof: Assume that ##f## is discontinuous at ##x_0##. That means for any sequence ##(x_n)##, in ##dom(f)##, converging to ##x_0## we don't have ## \lim f(x_n) = f(x_0)##, symbolically

For every N there exists, at least one, ##k > N## such that

$$

|x_k - x_0| < \delta ~ \text{but} ~ |f(x_k) f(x_0)| \geq \varepsilon

$$

Doubt 1: I haven't said anything about ##\delta## and ##\varepsilon## before, and simply introduced them to mean the difference is small,is this how we do it formally?

Now, as for every ##N## we have "a" ##k## satisfying those properties, therefore, we have all those ##k## forming a subset of ##\mathcal{N}## like thus

$$

k_1 > N_1 ~~\text{satisfying those properrties}$$

$$k_2 > N_2 > k_1 ~~ \text{satisfying those properties}$$

##\vdots ##

$$k_i > N_i > k_{i-1}$$

##\vdots##

and hence, ##(x_k)## is a sub-sequence of ##(x_n)##, and

$$\lim (x_k) \to x_0$$

##\since## for every ##x_k## we have ##|f(x_k) - f(x_0)| \geq \varepsilon##, but ##(x_k)## is a monotone sequence and we must have ##|f(x_k) - f(x_0)| < \varepsilon##

This contradiction has arisen because our assumption that ##f## was discontinuous was wrong.

Doubt #2: Is what I have done correct?

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